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
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