$\frac{1+\tan ^{2} x}{1-\tan ^{2} x} d x$ is equal to
- $\log \frac{1-\tan x}{1+\tan x}+c$
- $\log \frac{1+\tan x}{1-\tan x}+c$
- $\frac{1}{2} \log \frac{1-\tan x}{1+\tan x}+c$
- $\frac{1}{2} \log \frac{1+\tan x}{1-\tan x}+c$
The Correct Option is D
Solution and Explanation
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
$=\frac{\ln |\tan ( x )+1|}{2}-\frac{\ln |\tan ( x )-1|}{2}+ C$
$=\frac{1}{2} \log \frac{1+\tan x }{1-\tan x }+ C$
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
Method of Integration Using Partial Fractions:
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C