Question

$\int \frac{(xse{c}^{2}x+tanx)}{(xtanx+1)}dx=\frac{-x}{xtanx+1}f\left(x\right)+c$
then f(x)=

Answer 1

We have,

$\int \frac{(x{\sec }^{2}x+\tan x)}{(x\tan x+1)}dx=\frac{-x}{x\tan x+1}f\left(x\right)+C.......\left(1\right)$

find the value of $f\left(x\right)$.

Solve by L.H.S. and we get

Let,

$x\tan x+1=t$

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

$=\int \frac{dt}{t}$

$=\log t+C$

$=\log (x\tan x+1)+C$

Now put the value of equation (1) and we get,

$\Rightarrow \log (x\tan x+1)+C=\frac{-x}{x\tan x+1}f\left(x\right)+C$

$\Rightarrow \log (x\tan x+1)=\frac{-x}{x\tan x+1}f\left(x\right)$

$\Rightarrow f\left(x\right)=-\frac{(x\tan x+1)\log (x\tan x+1)}{x}$

$\Rightarrow f\left(x\right)=-\frac{\log {(x\tan x+1)}^{(x\tan x+1)}}{x}$

This is the value of $f\left(x\right)$

Hence, this is the answer.