Question

$\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx=f\left(x\right){\tan }^{-1}(g\left(x\right))-\frac{3}{4}\ln (h\left(x\right))+k$
If $h(-1)=9,$ then which of the following option(s) is/are correct?
$($ $k$ is a constant of integration $)$

• A. $f\left(x\right)=\frac{2}{3}(x+1)$
• B. $g\left(x\right)=\frac{2x+2}{3}$
• C. Minimum value of $h\left(x\right)$ is $9$
• D. $\frac{d}{dx}h\left(x\right)=8(x+1)$

Answer 1

Answer: (B), (C), (D)

$\frac{d}{dx}h\left(x\right)=8(x+1)$

$I=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$
$=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{(2x+2{)}^2+3^2}}\right)dx$

Put $2x+2=3\tan \theta ,\theta \in (\frac{-\pi }{2},\frac{\pi }{2})$
$\Rightarrow 2dx=3{\sec }^2\theta d\theta$
$I=\int {\sin }^{-1}\left(\frac{3\tan \theta }{3\sec \theta }\right)\frac{3}{2}{\sec }^2\theta d\theta$
$=\frac{3}{2}\int \theta {\sec }^2\theta d\theta$
$=\frac{3}{2}[\theta \tan \theta -\int \tan \theta d\theta ]$
$=\frac{3}{2}[\theta \tan \theta -\ln |\sec \theta |]+k_1$
$=\frac{3}{2}[\frac{2x+2}{3}{\tan }^{-1}\left(\frac{2x+2}{3}\right)-\ln \left(\sqrt{1+{\left(\frac{2x+2}{3}\right)}^2}\right)]+k_1$
$=\frac{3}{2}\left[\frac{2}{3}\right(x+1){\tan }^{-1}\left(\frac{2}{3}\right(x+1\left)\right)-\ln \sqrt{\frac{4x^2+8x+13}{9}}]+k_1,$
$\Rightarrow I=(x+1){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\ln \left(\frac{4x^2+8x+13}{9}\right)+k_1$
$\Rightarrow I=(x+1){\tan }^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\ln (4x^2+8x+13)+k$
where $k=k_1+\frac{3}{2}\ln 3$

$\Rightarrow f\left(x\right)=x+1$
$g\left(x\right)=\frac{2x+2}{3}$
$h\left(x\right)=\frac{4x^2+8x+13}{9}\text{or}4x^2+8x+13$
Since, $h(-1)=9$
$\therefore h\left(x\right)=4x^2+8x+13$
$=4(x+1{)}^2+9$
$\frac{d}{dx}h\left(x\right)=8(x+1)$