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

\frac{d ^{2}}{d x ^{2}} (x^{\frac{6}{7}} + 5 x)

y^{^{'}} = 1 + a y

\frac{d}{d x} (cos (5 x^{2} - 7 x) (10 x - 7))

x y^{^{'}} - 3 y = x^{3}

\int (- 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 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