What is the integral of (e^{6x})/(e^{12x)+49} ?

The integral of (e^{6x})/(e^{12x)+49} is 1/42 arctan(1/7 e^{6x})+C

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integral of (e^{6x})/(e^{12x)+49}

\int \frac{e ^{6 x}}{e ^{12 x} + 49} d x

\frac{1}{42} arctan (\frac{1}{7} e^{6 x}) + C

Solution steps

\int \frac{e ^{6 x}}{e ^{12 x} + 49} d x

Apply u-substitution

= \int \frac{e ^{u}}{6 ( e ^{2 u} + 49 )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{e ^{u}}{e ^{2 u} + 49} d u

Apply u-substitution

= \frac{1}{6} \cdot \int \frac{1}{v ^{2} + 49} d v

Apply Integral Substitution

= \frac{1}{6} \cdot \int \frac{1}{7 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{6} \cdot \frac{1}{7} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{6} \cdot \frac{1}{7} arctan (w)

Substitute back

= \frac{1}{6} \cdot \frac{1}{7} arctan (\frac{e ^{6 x}}{7})

Simplify

\frac{1}{6} \cdot \frac{1}{7} arctan (\frac{e ^{6 x}}{7}) : \frac{1}{42} arctan (\frac{1}{7} e^{6 x})

= \frac{1}{42} arctan (\frac{1}{7} e^{6 x})

Add a constant to the solution

= \frac{1}{42} arctan (\frac{1}{7} e^{6 x}) + C

What is the integral of (e^{6x})/(e^{12x)+49} ?
The integral of (e^{6x})/(e^{12x)+49} is 1/42 arctan(1/7 e^{6x})+C