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 $