What is the integral of (e^{2x})/(49+e^{4x)} ?

The integral of (e^{2x})/(49+e^{4x)} is 1/14 arctan(1/7 e^{2x})+C

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integral of (e^{2x})/(49+e^{4x)}

\int \frac{e ^{2 x}}{49 + e ^{4 x}} d x

\frac{1}{14} arctan (\frac{1}{7} e^{2 x}) + C

Solution steps

\int \frac{e ^{2 x}}{49 + e ^{4 x}} d x

Apply u-substitution

: \int \frac{1}{2 ( 49 + u ^{2} )} d u

= \int \frac{1}{2 ( 49 + u ^{2} )} d u

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{7} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{7} arctan (\frac{e ^{2 x}}{7})

Simplify

\frac{1}{2} \cdot \frac{1}{7} arctan (\frac{e ^{2 x}}{7}) : \frac{1}{14} arctan (\frac{1}{7} e^{2 x})

= \frac{1}{14} arctan (\frac{1}{7} e^{2 x})

Add a constant to the solution

= \frac{1}{14} arctan (\frac{1}{7} e^{2 x}) + C

(2 y - 9) d x + x d y = 0

\int \frac{1}{x ^{\frac{1}{3}} + x ^{\frac{1}{4}}} d x

75.6 x + \frac{36590.4}{x ^{2}}

\int \frac{1}{x ^{3} - x ^{2}} d x

What is the integral of (e^{2x})/(49+e^{4x)} ?
The integral of (e^{2x})/(49+e^{4x)} is 1/14 arctan(1/7 e^{2x})+C