Mathematics

# $\displaystyle \int \sqrt{x^{2}+2x+5}\; dx$ is equal to

$\displaystyle =\frac{1}{2}\left ( x+1 \right )\sqrt{x^{2}+2x+5}+2\:log$

$\displaystyle \left | \left ( x+1 \right )+\sqrt{x^{2}+2x+5} \right |+C$

##### SOLUTION
$\displaystyle I=\int \sqrt{x^{2}+2x+5}dx$
$\displaystyle =\int \sqrt{\left ( x+1 \right )^{2}+4}dx$
Put $x+1=t$
$\Rightarrow dx=dt$

$\displaystyle I=\int \sqrt{t^2+2^2}$
$\displaystyle =\frac { 1 }{ 2 } t\sqrt { t^{ 2 } +2^{ 2 } } +\frac { 1 }{ 2 } .\left( 2 \right) ^{ 2 }log\left| t+\sqrt { t^{ 2 }+2^{ 2 } } \right| +C$

$\displaystyle=\frac{1}{2}\left ( x+1 \right )\sqrt{\left ( x+1^{2} \right )+2^{2}} +\frac{1}{2}.\left ( 2 \right )^{2}log\left | \left ( x+1 \right ) +\sqrt{\left ( x+1 \right )^{2}+2^{2}}\right |+C$

$\displaystyle =\frac{1}{2}\left ( x+1 \right )\sqrt{x^{2}+2x+5}+2\:log \left | \left ( x+1 \right )+\sqrt{x^{2}+2x+5} \right |+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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