Question

$\int \frac{x^2(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx=$

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

Answer 1

Answer: (A)

$\frac{x^2}{x\tan x+1}-2log|x\sin x+\cos x|+C$

Let $I=\int x^2\cdot \left(\frac{x{\sec }^{2}x+\tan x}{x\tan x+1}\right)dx$

Using IBP & put $x\tan x+1=t$

then $(x{\sec }^{2}x+\tan x)dx=dt$

$I=x^2\int \frac{1}{t}dt-2\int \left(x\int \frac{1}{t}dt\right)dx$

$=\frac{x^2}{x\tan x+1}-2\int \frac{x}{x\tan x+1}dx$

$=\frac{x^2}{x\tan x+1}-2\int \frac{x\cos x}{x\sin x+\cos x}dx$

$\int \frac{x\cos x}{x\sin x+\cos x}dx$

$u=x\sin x+\cos x$

$x\cos xdx=du$

$=\frac{1}{u}du$

$=lnu$

$=ln(x\sin x+\cos x)$

$\int \frac{x\cos x}{x\sin x+\cos x}dx=ln|x\sin x+\cos x|+C$

$\frac{x^2}{x\tan x+1}-2$

$=\frac{x^2}{x\tan x+1}-2ln|x\sin x+\cos x|+C$.