Question

Solve $\sqrt{\dfrac{1+x}{1-x}}dx$.

Answer 1

To find the value of the integration we first take the denominator of the term given and then multiply it up and down or numerator and denominator to form two integrals which by integration substitution we get the value of anti-derivative but first we need to correctly multiply the denominator with numerator and denominator.

Complete step by step answer:
Using the method of simplifying fraction we multiply the numerator and denominator with the same value of $1-x$.
After multiplying the values in both numerator and denominator we get the new value of numerator and denominator as:
$\Rightarrow \int{\sqrt{\dfrac{1+x}{1-x}\times \dfrac{1+x}{1+x}}dx}$
$\Rightarrow \int{\sqrt{\dfrac{{{\left( 1+x \right)}^{2}}}{{{1}^{2}}-{{x}^{2}}}}dx}$
$\Rightarrow \int{\dfrac{1+x}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}$
After this we will separate the numerator into and with their respective but same denominator we get the two parts of the integral equation as:
$\Rightarrow \int{\dfrac{1}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}+\int{\dfrac{x}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}$
First we integrate the part and after that we place both of the parts together in addition with each other.
$\Rightarrow \int{\dfrac{1}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}+\int{\dfrac{x}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}$
The integrating value of $\int{\dfrac{1}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}$ is ${{\sin }^{-1}}x+C$.
And the integrating value of $\int{\dfrac{x}{\sqrt{{{1}^{2}}-{{x}^{2}}}}dx}$ is $\dfrac{1}{2}\left( -2\sqrt{1-{{x}^{2}}} \right)+C$.
Placing them in addition we get the value of the integral part as:
$\Rightarrow {{\sin }^{-1}}x+\dfrac{1}{2}\left( -2\sqrt{1-{{x}^{2}}} \right)+C$.

Therefore, the solution for the integration is ${{\sin }^{-1}}x+\dfrac{1}{2}\left( -2\sqrt{1-{{x}^{2}}} \right)+C$.

Note: In integral calculus, integration by substitution is a process in which integrals are evaluated and anti-derivatives are formed; it is also known as u-substitution or change of variables.