Question

If $I=\int si{n}^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx=(x+1)ta{n}^{-1}\frac{2x+2}{3}-\frac{A}{248}\log (4x^2+8x+13)+C$ then A is equal to.

Answer 1

$I=\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx=x{\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)-\int \frac{6x\sqrt{\frac{1}{4x^2+8x+13}}}{\sqrt{4x^2+8x+13}}dx=x{\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)-6\int \frac{x}{4x^2+8x+13}dx=x{\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)-\frac{3}{4}\int \frac{8x+8}{(4x^2+8x+13)}dx+6\int \frac{1}{4x^2+8x+13}dx=x{\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)+I_1+I_2$
Where
$I_1=-\frac{3}{4}\int \frac{8x+8}{(4x^2+8x+13)}dx$
Put $4x^2+8x+13=t\Rightarrow (8x+8)dx=dt$
$I_1=-\frac{3\log t}{4}=-\frac{3\log (4x^2+8x+13)}{4}$
And
$I_2=6\int \frac{1}{4x^2+8x+13}dx=6\int \frac{1}{{(2x+2)}^2+9}dx$
Put $u=2x+2\Rightarrow du=2dx$
$I_2=3\int \frac{1}{u^2+9}du={\tan }^{-1}\left(\frac{2x+2}{3}\right)$
Therefore,
$I=-\frac{3\log (4x^2+8x+13)}{4}+x{\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)+{\tan }^{-1}\left(\frac{2x+2}{3}\right)$
Hence $A=186$