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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`int ((1+tanx))/((x-logcosx))dx.` |
| Answer» Correct Answer - `-log |cos x+sin x|C` | |
| 2. |
Examples: ` int sec^2x / sqrt ( 16 + tan^2x) dx`A. `log|tanx+sqrt(tan^(2)x+16)|+C`B. `log|x+sqrt(tan^(2)x+16)|+C`C. `log|tan x-sqrt(tan^(2)x+16)|+C`D. None of these |
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Answer» Correct Answer - A Put `tan x=t and sec^(2)x dx =dt . `then , `I=int (dt)/(sqrt(t^(2)+4^(2)))=log |t+sqrt(t^(2)+16)|=log |tan x+sqrt(tan^(2)x+16)|+C.` |
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| 3. |
`int (x)/((x^(4)-16))dx=?`A. `(1)/(4)log |(x^(2)+4)/(x^(2)-4)+C`B. `(1)/(16)log |(x^(2)+4)/(x^(2)-4)+C`C. `(1)/(16)log |(x^(2)-4)/(x^(2)+4)+C`D. None of these |
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Answer» Correct Answer - C Put `x^(2)=t and 2xdx=dt.` `therefore I=(1)/(2)int(dt)/((t^(2)-4^(2)))=(1)/(2).(1)/(2xx4)log|(t-4)/(t+4)|=(1)/(16)log|(x^(2)-4)/(x^(2)+4)|+C.` |
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| 4. |
`int(x^(2))/(sqrt(x^(6)-1))dx=?`A. `(1)/(2) log |x^(3)+sqrt(x^(6)-1)|+C`B. `(1)/(3) log |x^(3)+sqrt(x^(6)-1)|+C`C. `(1)/(3) log |x^(3)-sqrt(x^(6)-1)|+C`D. None of these |
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Answer» Correct Answer - B Put `x^(3) = t and 3x^(2)dx=dt.` then `I=(1)/(3)int (dt)/(sqrt(t^(2)-1))=(1)/(3)log |t+sqrt(t^(2)-1)|+C=(1)/(3)log |x^(3)+sqrt(x^(6)-1)|+C.` |
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| 5. |
Evaluate:`int(sinx)/(sqrt(4cos^2x-1))dx`A. `-(1)/(2)log |2cosx+sqrt(4cos^(2)x-1)|+C`B. `-(1)/(3)log |2cosx+sqrt(4cos^(2)x-1)|+C`C. `-(1)/(6)log |cosx+sqrt(2cos^(2)x-1)|+C`D. None of these |
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Answer» Correct Answer - A Put ` cos x=t and sin x dx =-dt ,` then `I=-int(dt)/(sqrt(4t^(2)-1))=-(1)/(2)int (dt)/(sqrt(t^(2)-(1)/(4)))=-(1)/(2)log |t+sqrt(t^(2)-(1)/(4))|+C` `=-(1)/(2)log |2t+sqrt(4t^(2)-1)|+C=-(1)/(2) log|2cos x+sqrt(4 cos^(2)x-1)|+C.` |
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| 6. |
Evaluate:`int(3_(4x)-x^2)/((x+2)(x-1))dx` |
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Answer» Correct Answer - `-x+3log |x+2|+2log |x-1|+C` `((3+4x-x^(2)))/((x+2)(x-1))=((-x^(2)+4x+3))/((x^(2)+x-2))={-1+((5x+1))/((x+2)(x-1))}.` |
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| 7. |
Evaluate `int dx/(x(x^n+1))` |
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Answer» Putting `x^(n)=t,` We get `nx^(n-1)dx=dt.` `therefore (nx^(n))/(x)dx=dt implies(1)/(x)dx=(1)/(nt)dt("note").` `therefore int(dx)/(x(x^(n)+1))=int(dt)/(nt(t+1))=(1)/(n).int(dt)/(t(t+1)).` `Let (1)/(t(t+1))=(A)/(t)+(B)/((t+1)).` Then`,1-=A(t+1)+Bt.` Putting `t=0`in (i) ,we get `A=1.` Putting `t=-1` in (i), we get `B=-1.` `therefore (1)/(t(t+1))={(1)/(t)-(1)/((t+1))}` `=(1)/(n)[int(1)/(t)dt-int(1)/((t+1))dt]` `=(1)/(n).{log|t|-log|t+1|}+C` `=(1)/(n).log|(t)/(t|1)|+C` `=(1)/(n)log|(x^(n))/(x^(n)+1)|+C.` |
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| 8. |
Evaluate: `int(2x+5)/(x^2-x-2) dx` |
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Answer» Correct Answer - `3 log |x-2|-log |x-1|+C` `I=int((2x+5))/((x-2)(x+1))dx.` |
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| 9. |
Evaluate: `int(x^3)/((x-1)(x-2)) dx` |
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Answer» Correct Answer - `(x^(2))/(2)+3x-log |x-1|+8 log|x-2|+C` `(x^(3))/((x-1)(x+2))=(x^(3))/((x^(2)-3x+2))=[x+3+(7x-6)/(x^(2)-3x+2)]={x+3+(7x-6)/((x-2)(x-1))}.` |
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| 10. |
Evaluate :`int(x^2)/((x^2+4)(x^2+9))dx` |
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Answer» `Let (x^(2))/((x^(2)+4)(x^(2)+9))=(y)/((y+4)(y+9)),"where "x^(2)=y.` `Let (y)/((y+4)(y+9))=(A)/((y+4))+(B)/((y+9)).` then`,y-=A(y+9)+B(y+4).` Putting `y=-4`on both sides of(i) , we get `A=(-4)/(5).` Putting `y=-9`on both sides of (i) , we get `B=(9)/(5).` `therefore (y)/((y+4)(y+9))=(-4)/(5(y+4))+(9)/(5(y+9))` `implies(x^(2))/((x^(2)+4)(x^(2)+9))=(-4)/(5(x^(2)+4))+(9)/(5(x^(2)+9))` `implies int(x^(2))/((x^(2)+4)(x^(2)+9))dx=(-4)/(5)int(dx)/((x^(2)+4))+(9)/(5(x^(2)+9))=((-4)/(5)xx(1)/(2))tan^(-1)""(x)/(2)+((9)/(5)xx(1)/(3))tan^(-1)""(x)/(3)=(-2)/(5)tan^(-1)++(x)/(2)+(3)/(5)tan^(-1)""(x)/(3)+C.` |
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| 11. |
`int (dx)/((9+x^(2)))=?`A. `tan^(-1)""(x)/(3)+C`B. `(1)/(3)tan^(-1)""(x)/(3)+C`C. ` 3tan^(-1)""(x)/(3)+C`D. None of these |
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Answer» Correct Answer - B `I=int(dx)/((3^(2)+x^(2)))=(1)/(3)tan^(-1)""(x)/(3)+C.` |
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| 12. |
`int (dx)/((4+16x^(2)))=?`A. `(1)/(32)tan^(-1)4x+C`B. `(1)/(16)tan^(-1)""(x)/(2)+C`C. `(1)/(8)tan^(-1)""2x+C`D. `(1)/(4)tan^(-1)""(x)/(2)+C` |
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Answer» Correct Answer - C `I=int (1)/(16).int (dx)/(((1)/(4)+x^(2)))=(1)/(16).(dx)/({((1)/(2))^(2)+x^(2)})=(1)/(16).(1)/(((1)/(2)))tan^(-1)""(x)/(((1)/(2)))+X=(1)/(8)tan^(-1)2x+c.` |
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| 13. |
Evaluate `int (x^(2))/((1+x^(3))(2+x^(3)))dx.` |
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Answer» Putting `x^(3)=t and x^(2)dx=(1)/(3)dt,`we get `I=int(x^(2))/((1+x^(3))(2+x^(3)))dx=(1)/(3).int(dt)/((1+t)(2+t)).` `Let (1)/((1+t)(2+t))=(A)/((1+t))+(B)/((2+t)).`then, `1-=A(2+t)+B(1+t).` Putting `t=-1` in (i) , we get `A=1.` Putting `t=-2`in (i) , we get `B=-1.` `therefore (1)/((1+t)(2+t))=(1)/((1+t))-(1)/((2+t))` `implies I=int (dt)/((1+t)(2+t))=int(dt)/((1+t))-int(dt)/((2+t))` `=log |1+t|-log|2+t|+C` `=log |(1+t)/(2+t)|+C` `=log |(1+x^(3))/(2+x^(3))|+C` |
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| 14. |
Evaluate `int (dx)/(x{6(logx)^(2)+7logx+2}).` |
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Answer» Putting `log x=t and (1)/(x) dx=dt,` we get `I=int (dx)/(x{6(logx)^(2)+7logx+2})=int(dt)/((6t^(2)+7t+2))=int(dt)/((2t+1)(3t+2)).` `Let (1)/((2t+1)(3t+2))=(A)/((2t+1))+(B)/((3t+2)).` Then,`1-=A(3t+2)+B(2t+1).` Putting `t=-(1)/(2)`in (i) , we get `A=2` Putting `t=(-2)/(3)`in (i) , we get `B=-3.` `therefore (1)/((2t+1)(3t+2))=(2)/((2t+1))-(3)/((3t+2))` `implies I=int(dt)/((2t+1)(3t+2))` `=int (dt)/((2t+1)(3t+2))` `=log |2t+1|-log|3t+2|+C` `=log|(2t+1)/(3t+2)|+C` `=log |(2logx+1)/(3logx+2)|+C` |
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| 15. |
Evaluate `int(cos x)/((1-sinx)(2-sinx))dx.` |
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Answer» Putting sin `x=t` cos `x dx =dt,`we get `I=(cosx)/((1-sinx)(2-sinx))dx=int(dt)/((1-t)(2-t)).` `Let (1)/((1-t)(2-t))=(A)/((1-t))+(B)/((2-t))` `implies 1-=A(2-t)+B(1-t).` Putting `t=1` in (i) , we get `A=1. . . . (i).` Putting `t=2`in (i) , we get `B=-1.` `therefore (1)/((1-t)(2-t))=(1)/((1-t))-(1)/((2-t))` `implies int (cosx)/((-sinx)(2-sinx))dx=int(dt)/((1-t)(2-t))` `=int {(1)/((1-t))-(1)/((2-t))}dt` `=int (dt)/((1-t))-int(dt)/((2-t))` `=-log|1-t|+log|2-t|+C` `=log |(2-t)/(1-t)|+C=log |(2-sinx)/(1-sinx)|+C.` |
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| 16. |
`int((5x^(2)-18x+17))/((x-1)^(2)(2x-3))dx` |
| Answer» Correct Answer - `(4)/((x-1))+(5)/(2)log|2x-3|+C` | |
| 17. |
Evaluate `int(dx)/((e^(x)-1)).` |
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Answer» Putting `e^(x)=t and e^(x)dx=dt,i.e., dx=(1)/(t),`we get `I=int(dx)/((e^(x)-1))=int(dt)/(t(t-1)).` `Let(1)/(t(t-1))=(A)/(t)+(B)/(t(t-1)).` `Then,1-=A(t-1)+Bt.` Putting `t=0`in (i) ,we get `A=-1.` Putting `t=1`in (i) , we get `B=1.` `therefore (1)/(t(t-1))=(-1)/(t)+(1)/((t-1)).` `"Hence",I=int(dx)/((e^(x)-1))` `=int(dt)/(t(t-1))=int(-1)/(t)dt+int(1)/((t-1))dt` `=-log |t|+log|t-1|+C` `=log |(t-1)/(t)|+C` `=log |(e^(x)-1)/(e^(x))|+C` |
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| 18. |
Evaluate:`int(tantheta+tan^3theta)/(1+tan^3theta)dtheta` |
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Answer» WE have `((tan theta+tan^(3) theta))/((1+tan^(3) theta))=(tan theta(1+tan^(2) theta))/((1+tan^(3)theta))=(tan theta sec^(2)theta)/((1+tan^(3) theta)).` `therefore I=int((tan theta+ tan ^(3) theta))/((1+tan^(3) theta))` `= int (tan theta sec^(2) theta)/((1+tan^(3) theta))d theta` `=int (t)/((1+t^(3)))dt=int (t)/((1+t)(1-t+t^(2))dt,`where `tan theta =t.` `Let (t)/((1+t)(1-t+t^(2)))=(A)/((1+t))+((Bt+C))/((1-t+t^(2))).`then `t-=A(1-t+t^(2))+(Bt+C)(1+t).` Putting `t=-1` on both sides of (i) , we get `A=(-1)/(3).` Comparing the coefficients of `t^(2)` on both sides of (i) , we get `A+B=0impliesB=-A=(1)/(3).` Comparing the constant terms on both sides of (i) , we get `A+C=0impliesC=-A=(1)/(3).` `therefore (t)/((1+t)(1-t+t^(2)))=(-1)/(3(1+t))+(((1)/(3)t+(1)/(3)))/((1-t+t^(2))).` `Now,I=int(t)/((1+t)(1-t+t^(2)))dt` `=-(1)/(3)int(dt )/((1+t))+(1)/(6)int(2t)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))` `=-(1)/(3)int(dt )/((1+t))+(1)/(6)int((2t-1)+1)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))` `=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+1|+(1)/(2)int(dt)/((t^(2)-t+(1)/(4))+(3)/(4))` `=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+1|+(1)/(2)int (dt)/((t-1//2)^(2)+(sqrt(3)//2)^(2))` `=-(1)/(3)log|1+t|+(1)/(6)log|t^(2)+t+1|+(1)/(2).(2)/(sqrt(3))tan^(-1)((t-(1)/(2)))/((sqrt(3)//2))+C` `=-(1)/(3)log |1+t|+(1)/(6)log|t^(2)-t+1|+(1)/(sqrt(3))tan^(-1)((2t-1)/(sqrt(3)))+C` `=-(1)/(3) log |1+ tan theta|+(1)/(6)log |tan^(2) theta - tan theta +1|+(1)/(sqrt(3))tan^(-1)((2 tan theta -1)/(sqrt(3)))+C.` |
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| 19. |
Evaluate: `int((x^2+1)(x^2+2))/((x^2+3)(x^2+4)) dx` |
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Answer» We have `((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))=((t+1)(t+2))/((t+3)(t+4)),"where "x^(2)=t` `=((t^(2)+3t+2))/((t^(2)+7t+12))=1-((4t+10))/((t+3)(t+4)).` `Let ((4t+10))/((t+3)(t+4))+(A)/((t+3))+(B)/((t+4))` `implies (4t+10)-=A(t+4)+B(t+3).. . . (i)` Putting `t=-3`in(i) ,we get `A=-2` Putting `t=-4`in(i) ,we get `B=6`. `therefore ((4t+10))/((t+3)(t+4))=(-2)/((t+3))+(6)/((t+4)).` `thus ,((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4)=((t+1)(t+2))/((t+3)(t+4)),"where"x^(2)=t` `=((t^(2)+3t+2))/(t^(2)+7t+12))+1-((4t+10))/((t+3)(t+4))` `=1-{(-2)/((t+3))+(6)/((t+4))}["from(ii)"]` `={1+(2)/((t+3))-(6)/((t+4))}` `={1+(2)/((x^(2)+3))-(6)/((x^(2)+4))}.` `thereforeint((x^(2)+1)(x^(2)+2))/((x^(2)+3)(x^(2)+4))dx=int{1+(2)/((x^(2)+3))-(6)/((x^(2)+4))}dx.` `intdx+2int(dx)/((x^(2)+3))-6int(dx)/((x^(2)+4))` `=x+(2)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))-(6)/(2)tan^(-1)((x)/(2))+C` `=x+(2)/(sqrt(3))tan^(-1)((x)/(sqrt(3)))-3tan^(-1)((x)/(2))+C.` |
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| 20. |
Evaluate: `int(x^2+5x+3)/(x^2+3x+2) dx` |
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Answer» Correct Answer - `x+3 log |x+2|- log |x+1|+C` `I=int{1+((2x+1))/((x+2)(x+1))}dx.` |
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| 21. |
`int(x^(3))/((x^(2)-4))dx` |
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Answer» Correct Answer - `(x^(2))/(2)+2 log |x^(2)-4|+C` `I=int{1+(4x)/((x^(2)-4))}dx=int{1+(2)/((x-2)(x+2))}dx.` |
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| 22. |
Evaluate: `int(x^2)/(1+x^3) dx` |
| Answer» Correct Answer - `(1)/(3)log |1+x^(3)|+C` | |
| 23. |
Evaluate `int(dx)/((sinx-sin2x)).` |
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Answer» `int(dx)/((sinx-sin2x))=int (dx)/((sin s-2 sin x cos x))` `=int (dx)/(sinx(1-2cos x ))=int(sinx)/(sin^(2)x(1-2 cos x))dx` `intj (sin x)/((1-cos^(2) x)(1-2cos x))dx` `=-int(dt)/((1-t^(2))(1-2t)),`where cos x=t `= int (dt)/((t-1)(t+1)(1-2t)).` `Let (1)/((t-1)(t+1)(1-2t))=(A)/((t-1))+(B)/((t+1))+( C)/((1-2t)).` `then , 1-=A(t+1)(1-2t)+B(t-1)(1-2t)+C(t-1)(t+1).` Putting `t=1` in (i) , we get `A=(-1)/(2)` Putting `t=-1` in (ii), we get `B=(-1)/(6).` Putting `t=(1)/(2)` in (ii), we get `C=(-4)/(3).` `therefore I=-(1)/(2)int(dt )/((t-1))-(1)/(6). int (dt) /((t+1))-(4)/(3).int (dt)/((1-2t))` `=-(1)/(2)log|t-1|-(1)/(6)log |t+1|+(2)/(3).int(-2dt)/((1-2t))` `=-(1)/(2) log |t-1|-(1)/(6)log |t+1|+(2)/(3)log|1-2t+C` `=-(1)/(2) log |cos x-1|-(1)/(6)log |cos x+1|+(2)/(3)log |1-2 cos x|+c.` |
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| 24. |
Evaluate:`int(sin2x)/((1+sinx)2+sinx)dx` |
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Answer» Correct Answer - `-2log |1+sin x|+4log |2+sin x|+C` write `sin 2x=2 and x cos x. `put `sin x=t and cos x dx=dt.` |
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| 25. |
Evaluate:`int(x^4+1)/(x^2+1)dx` |
| Answer» Correct Answer - `(x^(3))/(3)-x+2 tan^(-1)x+C` | |
| 26. |
Evaluate: `int(x^4 dx)/((x-1)(x^2+1))` |
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Answer» `(x^(4))/((x-1)(x^(2)+1))=(x^(4))/((x^(3)-x^(2)+x-1))=(x+1)+(1)/((x^(3)-x^(2)+x-1))` `implies(x^(4))/((x-1)(x^(2)+1))=(x+1)+(1)/((x-1)(x^(2)+1)).` `Let (1)/((x-1)(x^(2)+1))=(A)/((x-1))+(bx+C)/((x^(2)+1)).then`, `1-=A(x^(2)+1)+(Bx+C)(x-1).` Putting `x=1` in (ii) , we get `A=(1)/(2).` Comparing the coefficients of `x^(2)` on both sides of (ii) , we get `A+B=0implies B=-A=-(1)/(2).` Comparing the constant Terms on both sides on (ii) , we get `A-C=1implies C=(A-1)=((1)/(2)-1)=(-1)/(2).` `therefore (1)/((x-1)(x^(2)+1))=(1)/(2(x-1))+(-(1)/(2)x-(1)/(2))/((x^(2)+1))` `therefore (x^(4))/((x-1)(x^(2)+1))=(x+1)+(1)/(2(x-1))-(1)/(2).((x+1))/((x^(2)+1))` `=int (x^(4))/((x-1)(x^(2)+1))dx=int (x+1)dx=int(x-1)dx+(1)/(2)int(dx)/((x-1))-(1)/(4).int (2x)/((x^(2)+1))dx-(1)/(2)int(dx)/((x^(2)+1))` `=(x^(2))/(2)+x+(1)/(2)log|x-1|-(1)/(4)log(x^(2)+1)-(1)/(2)tan^(-1)x+C.` |
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| 27. |
Evaluate: `int(3x-2)/((x+1)^2(x+3)) dx` |
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Answer» `Let ((3x-2))/((x+1)^(2)(x+3))=(A)/((x+1))+(B)/((x+1)^(2))+(C)/((x+3))` `implies (3x-2)-=A(x+1)(x+3)+B(x+3)+C(x+1)^(2).` Putting `x=-3` on both sides of (i) , we get `C=(-11)/(4).` Putting `x=-1` on both sides of (i), we get `B=(-5)/(2).` Comparing the coefficients of `x^(2)` on both sides of (i) ,we get `A+C=0impliesA=-C=(11)/(4).` `therefore ((3x-2))/((x+1)^(2)(x+3))=(11)/(4(x+1))-(5)/(2(x+1)^(2))-(11)/(4(x+3))` `implies int((3x-2))/((x+1)^(2)(x+3))dx=(11)/(4).int (dx)/((x+1))-(5)/(2).int (1)/((x+1)^(2))dx-(11)/(4).int(dx)/((x+3))` `=(11)/(4).log |x+1|+(5)/(2(x+1))-(11)/(4).log|x+3|+C` `=(11)/(4)/log|(x+1)/(x+3)|+(5)/(2(x+1))+C.` |
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| 28. |
Evaluate:`int(x^2+1)/(x^2-5x+6)dx` |
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Answer» Here the integrand is not a proper rational function , on dividing `(x^(2)+1)"by" (x^(2)-5x+6),`we get `((x^(2)+1))/((x^(2)-5x+6))=1+((5x-5))/((x^(2)-5x+6))=1+((5x-5))/((x-2)(x-3)).` `Now ,Let ((5x-5))/((x-2)(x-3))=(A)/((x-2))+(B)/((x-3))` `implies ((5x-5))/((x-2)(x-3))=(A(x-3)+B(x-2))/((x-2)(x-3))` `implies (5x-5-=A(x-3)+B(x-2).` Putting `x=2` on both sides of (i) we get `A=-5.` Putting `x=3` on both sides of (i) , we get `B=10.` `therefore ((x^(2)+1))/((x^(2)-5x+6))=1-(5)/((x-2))+(10)/((x-3))` `implies int((x^(2)+1))/((x^(2)-5x+6))dx=intdx-5int(dx)/((x-2))+10int(dx)/((x-3))` `=x-5 log |x-2|+10 log |x-3|+c.` |
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| 29. |
Resolve `(2x+1)/((x-1)(x^(2)+1))` into partial fractions. |
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Answer» Equating the like powers of x on both sides of (i) , we get `A+B=0,C-B=2 and A-C=1` On solving these equations , we get `A=(3)/(2),B=(-3)/(2)and C=(1)/(2).` `therefore (2x+1)/((x-1)(x^(2)+1))=(3)/(2(x-1))+(((-3)/(2)x+(1)/(2)))/(x^(2)+1)=[(3)/(2(x-1))+((1-3x))/(2(x^(2)+1))].` |
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| 30. |
Resolve `(16)/((x-2)(x+2)^(2))` into partial fractions. |
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Answer» `Let (16)/((x-2)(x+2)^(2))=(A)/(x-2)+(B)/(x+2)+( C)/((x+2)^(2))` `or (16)/((x-2)(x+2)^(2))=(A(x+2)^(2)+B(x-2)(x+2)+C(x-2))/((x-2)(x+2)^(2))` `therefore 16-= A(x+2)^(2)+b(x-2)(x+2)+C(x-2) . . . .(i)` `or 16-=(A+B)x^(2)+(4A+C)x+(4A-4B-2C).. . . .(ii)` Putting `(x-2)=0 or x=2` in (i), we get `A=1.` Putting `(x+2)=0 or x=-2` in (i), we get `C=-4.` Comparing the coefficients of `x^(2)` on both sides of (ii) , we get `A+B=0 or B=-A=-1` Thus `A=1,B=-1 and C=-4.` `therefore (16)/((x-2)(x+2)^(2))=[(1)/((x-2))-(1)/((x+2))-(4)/((x+2)^(2))].` |
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| 31. |
Resolve `(x^(3)-2x^(2)-13x-12)/(x^(2)-3x-10)` into partial fractions. |
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Answer» On dividing we get `(x^(3)-2x^(2)-13x-12)/(x^(2)-3x-10)=(x+1)-(2)/((x^(2)-3x-10))` `Let (2)/((x^(2)-3x-10))=(2)/((x-5)(x+2))=(A)/(x-5)+(B)/(x+2)` then`, (2)/((x-5)(x+2))+(A(x+2)+B(x-5))/((x-5)(x+2))` or `2-= A(x+2)+B(x-5).` putting `(x-5)=0 or x=5 `in (ii) , we get `A=(2//7)` Putting `(x+2)=0 or x=-2`in (ii) , we get `B=(-2//7)` `therefore (2)/((x^(2)-3x-10))=(2)/(7(x-5))-(2)/(7(x+2)).` `"hence", (x^(3)-2x^(2)-13x-12)/(x^(2)-3x-10)=(x+1)-(2)/(7(x-5))+(2)/(7(x+2)).` |
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| 32. |
` int[3x+5]/[x^3-x^2-x+1].dx` |
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Answer» `(x^(3)-x^(2)-x-1)=x^(2)(x-1)-(x-1)-=(x-1)(x^(2)-1)=(x-1)^(2)(x+1).` `Let (3x+5)/((x^(3)-x^(2)-x-1))=(3x+5)/((x-1)^(2)(x+1))=(A)/((x-1))+(B)/((x-1)^(2))+( C)/((x+1))` `implies (3x+5)-=A(x-1)(x+1)+B(x+1)+C(x-1)^(2).` Putting `x=1` on both sides of (i) we get `B=4.` Putting `x=-1` on both sides of `(i)`, we get `C=(1)/(2).` Comparing the coefficient of `x^(2)` on both sides of (i) , we get `A+C=0implies A=-C=(-1)/(2)` `therefore ((3x+5))/((x^(3)+x^(2)+x+1))=(-1)/(2(x-1))+(4)/((x-1)^(2))+(1)/(2(x+1))` `implies ((3x+5))/((x^(3)-x^(2)-x+1))dx=-(1)/(2)int(dx)/((x-1))+4int(dx)/((x-1))^(2)+(1)/(2)int(dx)/((x+1))=-(1)/(2)log|x-1|-(4)/((x-1))+(1)/(2)log|x+1|+C.` |
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| 33. |
Evaluate`int (dx)/((x^(3)+x^(2)+x+1)).` |
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Answer» We have `(1)/((x^(3)+x^(2)+x+1))=(1)/(x^(2)(x+1)+(x+1))=(1)/((x+1)(x^(2)+1)).` `Let (1)/((x+1)(x^(2)+1))=(A)/((x+1))+(Bx+C))/((x^(2)+1))` `implies 1-=A(x^(2)+1)+(Bx+C)(x+1).. . . .(i)` Putting `x=-1`on both sides of (i),we get `A=(1)/(2).` Comparing the coefficients of `x^(2)` on both sides of (i) , we get `A+B=0impliesB=-A=(-1)/(2).` COmparing the coefficients of x on both sides of (i) , we get `B+C=0impliesC=-B=(1)/(2).` `therefore (1)/((x+1)(x^(2)+1))=(1)/(2(x+1))+((-1)/(2)x+(1)/(2))/(x^(2)+1)` `therefore int(dx)/((x^(3)+x^(2)+x+1))=int(dx)/((x+1)(x^(2)+1))` `=(1)/(2).int(dx)/((x+1))-(1)/(2)int(x)/((x^(2)+1))dx+(1)/(2)(dx)/((x^(2)+1))` `=(1)/(2).int(dx)/((x+1))-(1)/(4).int(2x)/((x^(2)+1))dx+(1)/(2)int(dx)/((x^(2)+1))` `=(1)/(2)log |x+1|-(1)/(4) log |x^(2)+1|+(1)/(2) tan^(-1)x+C.` |
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| 34. |
`int(cosx)/((1+sinx)(2+sin x))dx` |
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Answer» Correct Answer - `log|(1+sin x)/(2+sin x)|+C` Putting `sin x=t` and` cos x dx=dt,`we get `I=int(dt)/((1+t)(2+t)).` |
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| 35. |
`int ((3sinphi-2) cosphi)/(5-cos^2phi-4sinphi) d phi` |
| Answer» Correct Answer - `3 log |sin phi -2|-(4)/((sin phi-2))+C` | |
| 36. |
Evaluate:`int(1-sin2x)/(x+cos^2x)dx` |
| Answer» Correct Answer - `log |x+cos^(2)x|+C` | |
| 37. |
`int 2^(x) " dx "` |
| Answer» Correct Answer - `(2^x)/(log2)+C` | |
| 38. |
`int(dx)/(sqrt(4-9x^(2)))=?` |
| Answer» Correct Answer - `(1)/(3) sin ^(-1)((3x)/(2))+C` | |
| 39. |
`int(sin^-1(cosx)).dx` |
| Answer» Correct Answer - `(pix)/(2)-(x^(2))/(2)+C` | |
| 40. |
`int (2x+1)(sqrt(x^(2)+x+1))dx` |
| Answer» Correct Answer - `(2)/(3) (x^(2) +x+1)^(3//2)+C` | |
| 41. |
`int x cos x^(2)dx` |
| Answer» Correct Answer - `(1)/(2)sin x^(2)+C` | |
| 42. |
`int cos x cos 3x dx` |
| Answer» Correct Answer - `(sin 4x)/(8)+(sin2x)/(4)+C` | |
| 43. |
`int sin 3x sin x dx` |
| Answer» Correct Answer - `(sin 2x)/(4)+(sin4x)/(8)+C` | |
| 44. |
`int (dx)/((sqrt(x+2)+sqrt(x+1)))` |
| Answer» Correct Answer - `(2)/(3)(x+2)^(3//2)-(2)/(3)(x+1)^(3//2)+C` | |