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

The integral of 1/(e^xsqrt(4+e^{2x))} is -(sqrt(4+e^{2x)})/(4e^x)+C

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

\int \frac{1}{e ^{x} 4 + e ^{2 x}} d x

- \frac{4 + e ^{2 x}}{4 e ^{x}} + C

Solution steps

\int \frac{1}{e ^{x} 4 + e ^{2 x}} d x

Apply u-substitution

: \int \frac{1}{u ^{2} 4 + u ^{2}} d u

= \int \frac{1}{u ^{2} 4 + u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{4 tan ^{2} ( v )} d v

= \int \frac{sec ( v )}{4 tan ^{2} ( v )} d v

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{sec ( v )}{tan ^{2} ( v )} d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( v )}{sin ^{2} ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= \frac{1}{4} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= \frac{1}{4} (- \frac{1}{w})

Substitute back

= \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} e ^{x} ) )})

Simplify

\frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} e ^{x} ) )}) : - \frac{4 + e ^{2 x}}{4 e ^{x}}

= - \frac{4 + e ^{2 x}}{4 e ^{x}}

Add a constant to the solution

= - \frac{4 + e ^{2 x}}{4 e ^{x}} + C

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