A) \[{{\sin }^{-1}}(x+1)+c\]
B) \[\sin \,\,{{h}^{-1}}(x+1)+c\]
C) \[\tan \,\,{{h}^{-1}}(x+1)+c\]
D) \[{{\tan }^{-1}}(x+1)+c\]
Correct Answer: D
Solution :
We have, \[I=\int{\frac{d\,x}{{{x}^{2}}+2x+2}}=\int{\frac{d\,x}{{{(x+1)}^{2}}+1}}\] \[\therefore \] \[\int{{{\tan }^{-1}}x\,dx=\frac{1}{1+{{x}^{2}}}+c}\] \[\therefore \] \[I=\int{\frac{dx}{1+{{(x+1)}^{2}}}={{\tan }^{-1}}(x+1)+c}\]
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