Evaluate the following intregals: x+1 x^2+1 dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{x}+1}{\sqrt{\text{x}^2+1}}\text{dx}$
let $\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+1)+\mu$
$\text{x}+1=\lambda(2\text{x})+\mu$
Compairing the coefficient of like powewrs of x,
$2\lambda=1\ \Rightarrow\lambda=\frac{1}{2}$
$\Rightarrow\mu=1$
So, $\text{I}=\int\frac{\frac{1}{2}(2\text{x})+1}{\sqrt{\text{x}^2+1}}\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x})}{\sqrt{\text{x}^2+1}}\text{dx}+\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}$
$\text{I}=\frac{1}{2}\times2\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$ $\big[\text{since}, \int\frac{1}{\sqrt{\text{x}}}\text{dx}=2\sqrt{\text{x}}+\text{c},\int\frac{1}{\sqrt{\text{x}^2+1}}\text{dx}=\log\big|\text{x}+\sqrt{\text{x}^2-\text{x}^2}\big|+\text{C}\big]$
$\text{I}=\sqrt{\text{x}^2+1}+\log\big|\text{x}+\sqrt{\text{x}^2+1}\big|+\text{C}$

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