Question

$\int \frac{{\sec }^{2}xdx}{(1+\tan x)(3+\tan x)}$.

• A. $\frac{1}{2}ln\left(\frac{2+\tan x}{3+\tan x}\right)+C$
• B. $\frac{1}{2}ln\left(\frac{3+\tan x}{2+\tan x}\right)+C$
• C. $\frac{1}{2}ln\left(\frac{1+\tan x}{3+\tan x}\right)+C$
• D. $\frac{1}{2}ln\left(\frac{1+\tan x}{2+\tan x}\right)+C$

Answer 1

The correct option is C $\frac{1}{2}ln\left(\frac{1+\tan x}{3+\tan x}\right)+C$

$\int \frac{{\sec }^{2}xdx}{(1+\tan x)(3+\tan x)}$

Put $\tan x=t$

${\sec }^{2}dx\cdot dt$

$\int \frac{dt}{(1+b)(3+t)}\cdot \int \frac{A}{1+t}+\frac{B}{3+t}dt$

$3A+B=1$

$A+B=0$

$=\frac{1}{2}\int \frac{dt}{1+t}-\frac{1}{2}\int \frac{dt}{3+t}$

$3A-A=1$

$A=\frac{1}{2},B=-\frac{1}{2}$

$\frac{1}{2}ln(1+t)-\frac{1}{2}ln(3+t)$

$=\frac{1}{2}ln\left(\frac{1+\tan x}{3+\tan x}\right)+c$.