A) \[x-2{{\tan }^{-1}}(x/2)+c\]
B) \[x+2{{\tan }^{-1}}(x/2)+c\]
C) \[x-4{{\tan }^{-1}}(x/2)+c\]
D) \[x+4{{\tan }^{-1}}(x/2)+c\]
Correct Answer: A
Solution :
\[I=\int{\frac{{{x}^{2}}}{{{x}^{2}}+4}dx}\]\[=\int{\frac{{{x}^{2}}+4-4}{({{x}^{2}}+4)}dx}\] Þ \[I=\int{\left[ 1-\frac{4}{{{x}^{2}}+4} \right]\,dx}\]\[=\int{dx-\int{\frac{4}{{{x}^{2}}+4}dx}}\] \[\Rightarrow I=x-4\int{\frac{dx}{{{x}^{2}}+{{(2)}^{2}}}}\]\[=x-\frac{4}{2}{{\tan }^{-1}}(x/2)+c\] \[=x-2{{\tan }^{-1}}\frac{x}{2}+c\].
You need to login to perform this action.
You will be redirected in
3 sec
