What is the integral of (e^x)/((6e^x+1)^{3/2)} ?

The integral of (e^x)/((6e^x+1)^{3/2)} is -1/(3(6e^x+1)^{1/2)}+C

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integral of (e^x)/((6e^x+1)^{3/2)}

\int \frac{e ^{x}}{( 6 e ^{x} + 1 ) ^{\frac{3}{2}}} d x

- \frac{1}{3 ( 6 e ^{x} + 1 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{e ^{x}}{( 6 e ^{x} + 1 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( 6 u + 1 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( 6 u + 1 ) ^{\frac{3}{2}}} d u

Apply u-substitution

: \int \frac{1}{6 v ^{\frac{3}{2}}} d v

= \int \frac{1}{6 v ^{\frac{3}{2}}} d v

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: - \frac{2}{v ^{\frac{1}{2}}}

= \frac{1}{6} (- \frac{2}{v ^{\frac{1}{2}}})

Substitute back

= \frac{1}{6} (- \frac{2}{( 6 e ^{x} + 1 ) ^{\frac{1}{2}}})

Simplify

\frac{1}{6} (- \frac{2}{( 6 e ^{x} + 1 ) ^{\frac{1}{2}}}) : - \frac{1}{3 ( 6 e ^{x} + 1 ) ^{\frac{1}{2}}}

= - \frac{1}{3 ( 6 e ^{x} + 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{1}{3 ( 6 e ^{x} + 1 ) ^{\frac{1}{2}}} + C

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What is the integral of (e^x)/((6e^x+1)^{3/2)} ?
The integral of (e^x)/((6e^x+1)^{3/2)} is -1/(3(6e^x+1)^{1/2)}+C