Question

$\int {\sin }^{-1}\left\{\frac{\left(2x+2\right)}{\sqrt{4x^2+8x+13}}\right\}dx$ is equal to

• A. $\left(x+1\right){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log \left(\frac{4x^2+8x+13}{9}\right)+c$
• B. $\frac{3}{2}{\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log \left(\frac{4x^2+8x+13}{9}\right)+c$
• C. $\left(x+1\right){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{2}\log \left(4x^2+8x+13\right)+c$
• D. $\frac{3}{2}\left(x+1\right){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log \left(4x^2+8x+13\right)+c$

Answer 1

Answer: (A)

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