Mathematics

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


SOLUTION
$$\int e^{tan^{-1}x}(\frac{1+x+x^{2}}{1+x^{2}})dx$$
put $$tan^{-1}x=t$$
x= tan t
$$dx = sec^{2}t$$ $$dt$$
$$\int e^{t}(\frac{1+ tan t+ tan^{2}t}{1+tan^{2}t})sec^{2}t dt$$
$$(sec^{2}x- tan^{2}x=1 \Rightarrow 1+ tan^{2}x = sec^{2}x)$$
$$\int e^{t}(\frac{tan t + sec^{2}t}{sec^{2}t})sec^{2}t dt$$
$$\int e^{t}(\frac{tan t sec^{2}t}{sec^{2}t}+\frac{sec^{2}t sec^{2}t}{sec^{2}t})dt$$
$$\int e^{t}(tan t + sec^{2}t)dt$$
$$\int e^{t} tan t dt + \int e^{t} sec^{2}t dt$$
Integrate by parts $$I_{1}$$
$$ tan t e^{t}- \int sec^{2}t \int e^{t}dt + \int e^{t} sec^{2} t dt$$
$$tan t e^{t} - \int sec^{2}t \int e^{t} dt + \int e^{t} sec^{2} t dt$$
$$= e^{t} tan^{t}+C$$
Now put $$t= tan^{-1}x$$
$$= e^{tan^{-1}x} tan x (tan^{-1}x)+C$$
$$= e^{tan^{-1}x}+C$$
$$= xe^{tan^{-1}x}+C$$
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Subjective Medium Published on 17th 09, 2020
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