Question

$\frac{1+\tan ^{2} x}{1-\tan ^{2} x} d x$ is equal to

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

Answer 1

Answer: (D)

$\frac{1}{2} \log \frac{1+\tan x}{1-\tan x}+c$

Answer 2

$\int \frac{1+\tan ^{2}( x )}{1-\tan ^{2}( x )} dx =\int \frac{\sec ^{2}( x )}{1-\tan ^{2}( x )} dx$
Apply $u-$ substitution $: u=\tan (x)$
$\int \frac{1}{1-u^{2}} d u$
Use the common integral :
$\int \frac{1}{1-u^{2}} d u=\frac{\ln |u+1|}{2}-\frac{\ln |u-1|}{2}$
$=\frac{\ln | u +1|}{2}-\frac{\ln | u -1|}{2}$
Substitute back $u=\tan (x)$
$=\frac{\ln |\tan ( x )+1|}{2}-\frac{\ln |\tan ( x )-1|}{2}$
Add a constant to the solution
$=\frac{\ln |\tan ( x )+1|}{2}-\frac{\ln |\tan ( x )-1|}{2}+ C$
$=\frac{1}{2} \log \frac{1+\tan x }{1-\tan x }+ C$