What is the integral of (e^{arctan(4x)})/(1+16x^2) ?

The integral of (e^{arctan(4x)})/(1+16x^2) is 1/4 e^{arctan(4x)}+C

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integral of (e^{arctan(4x)})/(1+16x^2)

\int \frac{e ^{a r c t a n (4 x)}}{1 + 16 x ^{2}} d x

\frac{1}{4} e^{a r c t a n (4 x)} + C

Solution steps

\int \frac{e ^{a r c t a n (4 x)}}{1 + 16 x ^{2}} d x

Apply u-substitution

: \int \frac{e ^{a r c t a n (u)}}{4 ( 1 + u ^{2} )} d u

= \int \frac{e ^{a r c t a n (u)}}{4 ( 1 + u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{e ^{a r c t a n (u)}}{1 + u ^{2}} d u

Apply u-substitution

: \int e^{v} d v

= \frac{1}{4} \cdot \int e^{v} d v

Use the common integral:

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

= \frac{1}{4} e^{v}

Substitute back

= \frac{1}{4} e^{a r c t a n (4 x)}

Add a constant to the solution

= \frac{1}{4} e^{a r c t a n (4 x)} + C

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What is the integral of (e^{arctan(4x)})/(1+16x^2) ?
The integral of (e^{arctan(4x)})/(1+16x^2) is 1/4 e^{arctan(4x)}+C