Mathematics

# $\displaystyle\int 5^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{x^2 + 1}\right)dx$

##### SOLUTION
We have, $\dfrac{d}{dx}(x+\tan^{-1}x)=1+\dfrac{1}{1+x^2}=\dfrac{x^2+2}{x^2+1}$........(1)
$\displaystyle\int 5^{x + \tan^{-1}x}\left(\dfrac{x^2 + 2}{x^2 + 1}\right)dx$
$=\displaystyle\int 5^{x + \tan^{-1}x}d\left(x+\ tan^{-1}x\right)$ [ Using (1)]
$=\dfrac{5^{x+\tan^{-1}x}}{\log 5}+c$ [ Where $c$ is integrating constant]

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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