Question

The value of the integral $\int \frac{x^2(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx$ is

• A. $\frac{-x^2}{x\tan x+1}+2\log |x\cos x+\sin x|+c$
• B. $\frac{-x^2}{x\tan x+1}+\log |x\cos x+\sin x|+c$
• C. $\frac{-x^2}{x\tan x+1}-2\log |x\sin x+\cos x|+c$
• D. $\frac{-x^2}{x\tan x+1}+2\log |x\sin x+\cos x|+c$

Answer 1

The correct option is D $\frac{-x^2}{x\tan x+1}+2\log |x\sin x+\cos x|+c$

$I=\int \frac{x^2(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx$
Put $x\tan x+1=t\Rightarrow (x{\sec }^{2}x+\tan x)dx=dt$
$\int \frac{(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx=\int \frac{dt}{t^2}=\frac{-1}{t}=\frac{-1}{x\tan x+1}$

Now, using integration by parts
$\Rightarrow I=x^2\int \frac{(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx-\int 2x(\int \frac{(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx)dx$
$\Rightarrow I=x^2\left(\frac{-1}{x\tan x+1}\right)+\int \frac{2x}{x\tan x+1}dx\Rightarrow I=x^2\left(\frac{-1}{x\tan x+1}\right)+\int \frac{2x\cos x}{x\sin x+\cos x}dx$

Put $x\sin x+\cos x=z\Rightarrow x\cos xdx=dz$
$\Rightarrow I=\left(\frac{-x^2}{x\tan x+1}\right)+2\log |x\sin x+\cos x|+c$