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

x \to 0 lim (\frac{5 sin ( 2 x )}{x})

t e^{5 t}

w \to - 2 lim (\frac{w + 2}{w ^{2} - 6 w - 16})

\int tan (t) d t

\int 3 θ cos (\frac{θ}{5}) d θ

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