A) \[(x+1){{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\] \[-\frac{3}{4}\log \left( \frac{4{{x}^{2}}+8x+13}{9} \right)+c\]
B) \[\frac{3}{2}{{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\] \[-\frac{3}{4}\log \left( \frac{4{{x}^{2}}+8x+13}{9} \right)+c\]
C) \[(x+1){{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\] \[-\frac{3}{2}\log \,(4{{x}^{2}}+8x+13)+c\]
D) \[\frac{3}{2}(x+1){{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\] \[-\frac{3}{4}\log \,(4{{x}^{2}}+8x+13)+c\]
Correct Answer: A
Solution :
Let \[I=\int{{{\sin }^{-1}}}\left\{ \frac{2x+2}{\sqrt{4{{x}^{2}}+8x+13}} \right\}dx\] \[\Rightarrow \]\[I=\int{{{\sin }^{-1}}}\left\{ \frac{2x+2}{\sqrt{4{{x}^{2}}+8x+4+9}} \right\}dx\] \[\Rightarrow \]\[I=\int{{{\sin }^{-1}}}\left\{ \frac{2x+2}{\sqrt{{{(2x+2)}^{2}}+{{3}^{2}}}} \right\}dx\] Substituting \[2x+2=3\,\tan \theta ,\] \[\Rightarrow \]\[2dx=3{{\sec }^{2}}\theta d\theta ,\] we get \[I=\int{{{\sin }^{-1}}}\left\{ \frac{3\tan \theta }{3\sec \theta } \right\}\cdot \frac{3}{2}{{\sec }^{2}}\theta d\theta \] \[\Rightarrow \] \[I=\frac{3}{2}\int{{{\sin }^{-1}}}(\sin \theta )\cdot {{\sec }^{2}}\theta d\theta \] \[\Rightarrow \] \[I=\frac{3}{2}\int{\theta }se{{c}^{2}}\theta d\theta \] \[\Rightarrow \] \[I=\frac{3}{2}\int{[\theta }\tan \theta -\int{\,\tan \,}d\theta ]\] (integrating by parts) \[I=\frac{3}{2}[\theta \,tan\theta -log\left| \sec \theta \right|\,]+c\] \[=\frac{3}{2}\left[ {{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\cdot \left( \frac{2x+2}{3} \right) \right.\] \[\left. -\log \sqrt{1+{{\left( \frac{2x+2}{3} \right)}^{2}}} \right]+c\] \[=(x+1){{\tan }^{-1}}\left( \frac{2x+2}{3} \right)\] \[-\frac{3}{4}\log \left( \frac{4{{x}^{2}}+8x+13}{9} \right)+c\]
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