InterviewSolution
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InterviewSolution
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Evaluate: `int(3x-2) sqrt(x^2+x+1) dx` |
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Answer» Let `(3x-2)=A*(d)/(dx)(x^(2)+x+1)+B`. Then , `( 3x -2) =A (2x =1)+B`. Comparing the coefficients of like powers of x , we get `2A=3and A+B =-2. "So",A=(3)/(2)andB=(-7)/(2)`. `therefore(3x-2)=(3)/(2)(2x+1)-(7)/(2)*` So , `int(3x-2)sqrt(x^(2)+x+1)dx` `=int{(3)/(2)(2x+1)-(7)/(2)}sqrt(x^(2)+x+1)dx` `=(3)/(2)int(2x+1)sqrt(x^(2)+x+1)dx-(7)/(2)intsqrt(x^(2)+x+1)dx` `=(3)/(2)intsqrt(t)dt-(7)/(2)intsqrt((x^(2)+x+(1)/(4))+(3)/(4))dx`, where `x^(2)+x+1=t` in the 1st integral `=t^(3//2)-(7)/(2)intsqrt({(x+(1)/(2))^(2)+((sqrt(3))/(2))^(2)})dx` `=(x^(2)+x+1)^(3//2)-(7)/(2)*intsqrt(u^(2)+((sqrt(3))/(2))^(2))du, "where"u=(x+(1)/(2))` `=(x^(2)+x+1)^(3//2)-(7)/(2){(u)/(2)*sqrt(u^(2)+(3)/(4))+(3)/(8)log|u+sqrt(u^(2)+(3)/(4))|}+C` `{becauseintsqrt(u^(2)+a^(2))du=(u)/(2)sqrt(u^(2)+a^(2))+(a^(2))/(2)log|u+sqrt(u^(2)+a^(2))|+C}` `=(x^(2)+x+1)^(3//2)-(7u)/(8)sqrt(4u^(2)+3)-(21)/(16)log|u+sqrt(u^(2)+(3)/(4))|+C` `=(x^(2)+x+1)^(3//2)-(7)/(8)(x+(1)/(2))sqrt(4(x+(1)/(2))^(2)+3)-(21)/(16)log|(x+(1)/(2))+sqrt((x+(1)/(2))^(2)+(3)/(4))|+C` `=(x^(2)+x+1)^(3//2)-(7(2x+1))/(8)sqrt(x^(2)+x+1)-(21)/(16)log|((2x+1))/(2)+sqrt(x^(2)+x+1)|+C`. |
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