`int(x^(2)+1)/(x(x^(2)-1))dx=`A. `log((x^(2)-1)/(x))+c`B. `-log((x^(2)-1)/(x))+c`C. `log((x)/(x^(2)+1))+c`D. `-log((x)/(x^(2)+1))+c`

`int(x^(2)+1)/(x(x^(2)-1))dx=`A. `log((x^(2)-1)/(x))+c`B. `-log((x^(2)

1.	`int(x^(2)+1)/(x(x^(2)-1))dx=`A. `log((x^(2)-1)/(x))+c`B. `-log((x^(2)-1)/(x))+c`C. `log((x)/(x^(2)+1))+c`D. `-log((x)/(x^(2)+1))+c`
Answer» Correct Answer - A

`int2^(3x).3^(x)dx=`A. `2^(3x).3^(2x).(log2)(log3)+c`B. `2^(3x-1).3^(2x-1).(log2)(log3)+c`C. `2^(3x).3^(2x).log5+c`D. `2^(3x).3^(2x).log_(72)e+c`
`int(1-cosx)."cosec"^(2)xdx=`A. `tan((x)/(2))+c`B. `cot((x)/(2))+c`C. `(1)/(2)tan((x)/(2))+c`D. `2tan((x)/(2))+c`
`int sec^(8//9)x cosec^(10//9)x ` dx is equal toA. `-(cot^(1//9)x)+c`B. `9.(tan^(1//9)x)+c`C. `-9(cot^(1//9)x)+c`D. `-(1)/(9)(cot^(9)x)+c`
Evaluate: `int1/(sin^2x+sin2x) dx`A. `(1)/(2)log[(tanx)/(2+tanx)]+c`B. `log[sqrt((2+tanx)/(tanx))]+c`C. `(1)/(2)tan^(-1)[(tanx)/(2+tnax)]+c`D. `(sin2x)/(2)-(cos4x)/(4)+c`
If the integral `int(5tanx)/(tanx-2)dx = x+a"ln"|sinx-2cosx|+k`, then a is equal to:A. `-1`B. `-2`C. 1D. 2
`int(1)/(3+2cosx)dx=`A. `(2)/(5)tan^(-1)[(tan(x//2))/(5)]+c`B. `(5)/(sqrt2)tan^(-1)[(tan(x//2))/(sqrt2)]+c`C. `(sqrt2)/(5)tan^(-1)[(tan(x//2))/(5)]+c`D. `(2)/(sqrt5)tan^(-1)[(tan(x//2))/(sqrt5)]+c`
`int1/(5+3cosx)dx`A. `4log(tanx+5)+c`B. `4tan^(-1)(2tanx)+c`C. `(1)/(4)log((1)/(2)tanx)+c`D. `(1)/(4)tan^(-1)((1)/(2)tanx)+c`
`int(3)/(4-5sinx)dx=`A. `log[(tan(x//2)-2)/(tan(x//2)-1)]+c`B. `log[(2tanx-4)/(2tanx-1)]+c`C. `log[(2tan(x//2)-4)/(2tan(x//2)-1)]+c`D. `tan^(-1)(2tan.(x)/(2)-5)+c`
If `int(x+1)sqrt(2x-1)dx=a(2x-1)^(5//2)+b(2x-1)^(3//2)+c`, then `(b)/(a)=`A. 6B. 5C. 4D. 3
If `int(1)/(sqrt(1-2x))dx=a(1-2x)^(3//2)+bsqrt(1-2x)+c`, then `(b)/(a)=`A. `-3`B. `-2`C. 3D. 2