What is the integral of [(e^{arctan(x)}-2x)/(1+x^2)] ?

The integral of [(e^{arctan(x)}-2x)/(1+x^2)] is e^{arctan(x)}-ln|1+x^2|+C

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integral of [(e^{arctan(x)}-2x)/(1+x^2)]

\int [\frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}}] d x

e^{a r c t a n (x)} - ln 1 + x^{2} + C

Solution steps

\int \frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}} d x

Expand

\frac{e ^{a r c t a n (x)} - 2 x}{1 + x ^{2}} : \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} - \frac{2 x}{1 + x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x - \int \frac{2 x}{1 + x ^{2}} d x

\int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x = e^{a r c t a n (x)}

\int \frac{2 x}{1 + x ^{2}} d x = ln 1 + x^{2}

= e^{a r c t a n (x)} - ln 1 + x^{2}

Add a constant to the solution

= e^{a r c t a n (x)} - ln 1 + x^{2} + C

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What is the integral of [(e^{arctan(x)}-2x)/(1+x^2)] ?
The integral of [(e^{arctan(x)}-2x)/(1+x^2)] is e^{arctan(x)}-ln|1+x^2|+C