Q63 / 120 · VITEEE 2010 · Mathematics
$\int \sin ^{-1}\left\{\frac{(2 x+2)}{\sqrt{4 x^{2}+8 x+13}}\right\} d x$ is equal to
- Correct answerA$(x+1) \tan ^{-1}\left(\frac{2 x+2}{3}\right)-\frac{3}{4} \log \left(\frac{4 x^{2}+8 x+13}{9}\right)+c$
- B$\frac{3}{2} \tan ^{-1}\left(\frac{2 x+2}{3}\right)-\frac{3}{4} \log \left(\frac{4 x^{2}+8 x+13}{9}\right)+c$
- C$(x+1) \tan ^{-1}\left(\frac{2 x+2}{3}\right)-\frac{3}{2} \log \left(4 x^{2}+8 x+13\right)+c$
- D$\frac{3}{2}(x+1) \tan ^{-1}\left(\frac{2 x+2}{3}\right)-\frac{3}{4} \log \left(4 x^{2}+8 x+13\right)+c$
Solution
$$
\text { (a) Let } \begin{array}{l}
I=\int \sin ^{-1}\left\{\frac{2 x+2}{\sqrt{4 x^{2}+8 x+13}}\right\} d x \\
\Rightarrow I=\int \sin ^{-1}\left\{\frac{2 x+2}{\sqrt{4 x^{2}+8 x+4+9}}\right\} d x \\
\Rightarrow I=\int \sin ^{-1}\left\{\frac{2 x+2}{\sqrt{(2 x+2)^{2}+3^{2}}}\right\} d x
\end{array}
$$
$$
\text { Substituting } 2 x+2=3 \tan \theta \text {, }
$$
$$
\begin{array}{l}
\Rightarrow 2 d x=3 \sec ^{2} \theta d \theta, \text { we get } \\
I=\int \sin ^{-1}\left\{\frac{3 \tan \theta}{3 \sec \theta}\right\} \cdot \frac{3}{2} \sec ^{2} \theta d \theta \\
\Rightarrow I=\frac{3}{2} \int \sin ^{-1}(\sin \theta) \cdot \sec ^{2} \theta d \theta \\
\Rightarrow I=\frac{3}{2} \int \theta \sec ^{2} \theta d \theta \\
\Rightarrow I=\frac{3}{2}\left[\theta \tan \theta-\int \tan \theta d \theta\right]
\end{array}
$$
(integrating by parts)
$$
\begin{aligned}
I &=\frac{3}{2}[\theta \tan \theta-\log |\sec \theta|]+c \\
&=\frac{3}{2}\left[\tan ^{-1}\left(\frac{2 x+2}{3}\right) \cdot\left(\frac{2 x+2}{3}\right)\right.\\
&\left.\quad-\log \sqrt{1+\left(\frac{2 x+2}{3}\right)^{2}}\right]+c \\
&=(x+1) \tan ^{-1}\left(\frac{2 x+2}{3}\right) \\
&-\frac{3}{4} \log \left(\frac{4 x^{2}+8 x+13}{9}\right)+c
\end{aligned}
$$
Hence, the correct answer is option: 'A'.