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

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

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

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

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

Solution steps

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

(\frac{3 y ^{2} - t ^{2}}{y ^{5}}) \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

\frac{\partial}{\partial x} (x e^{- y})

\int (sec (2 y) - csc (2 y)) d y

x \to - 1 lim (1)

\int (x - 6)^{3} d x

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