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

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

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

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

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

Solution steps

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

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 (y)

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

Add a constant to the solution

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

\int x arccos (x) d x

\int (e^{2 x} cos (3 x)) d x

3 x^{2} - 6 x + 5

\int \frac{cos ( x )}{1 + 2 sin ( x )} d x

\frac{d}{d x} (e^{l n (x^{2})})

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