A) \[{{\tan }^{-1}}(x+1)\]
B) \[2{{\tan }^{-1}}(x+1)\]
C) \[-{{\tan }^{-1}}(x+1)\]
D) \[3{{\tan }^{-1}}(x+1)\]
Correct Answer: A
Solution :
Let \[I=\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+2x+2}}\] \[=\int_{{}}^{{}}{\frac{dx}{1+{{(x+1)}^{2}}}}\] \[={{\tan }^{-1}}(x+1)+c\] But \[I=f(x)+c\] \[\therefore \] \[f(x)={{\tan }^{-1}}(x+1)\]
You need to login to perform this action.
You will be redirected in
3 sec
