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

The integral of (e^{arctan(x)})/(x^2+1) is e^{arctan(x)}+C

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

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

e^{a r c t a n (x)} + C

Solution steps

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

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e^{u}

Substitute back

u = arctan (x)

= e^{a r c t a n (x)}

Add a constant to the solution

= e^{a r c t a n (x)} + C

\int \frac{3 - 4 x}{16 - 9 x ^{2}} d x

f (x) = \frac{x - 1}{x + 1}

\int e^{x} \cdot arctan (e^{x}) d x

f (x) = x^{2} - 4, (5, 21)

\int (e^{7 x} - \frac{1}{x}) d x

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