What is the integral of 8/(sqrt(e^x-2)) ?

The integral of 8/(sqrt(e^x-2)) is 8sqrt(2)arctan(sqrt(1/2 (e^x-2)))+C

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integral of 8/(sqrt(e^x-2))

\int \frac{8}{e ^{x} - 2} d x

82 arctan (\frac{1}{2} (e^{x} - 2)) + C

Solution steps

\int \frac{8}{e ^{x} - 2} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{e ^{x} - 2} d x

Apply u-substitution

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

= 8 \cdot \int \frac{2}{u ^{2} + 2} d u

Take the constant out:

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

= 8 \cdot 2 \cdot \int \frac{1}{u ^{2} + 2} d u

Apply Integral Substitution

= 8 \cdot 2 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 8 \cdot 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 8 \cdot 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 8 \cdot 2 \cdot \frac{1}{2} arctan (\frac{e ^{x} - 2}{2})

Simplify

8 \cdot 2 \cdot \frac{1}{2} arctan (\frac{e ^{x} - 2}{2}) : 82 arctan (\frac{1}{2} (e^{x} - 2))

= 82 arctan (\frac{1}{2} (e^{x} - 2))

Add a constant to the solution

= 82 arctan (\frac{1}{2} (e^{x} - 2)) + C

\int_{- \infty}^{15} r e^{\frac{r}{5}} d r

\frac{d}{d x} (3 x^{2} + 8)

\int_{- 1}^{2} (3 x^{2} + 2) d x

x \to \frac{1}{3} x lim (sin (x))

y^{^{'}} = e^{2 x - y}

What is the integral of 8/(sqrt(e^x-2)) ?
The integral of 8/(sqrt(e^x-2)) is 8sqrt(2)arctan(sqrt(1/2 (e^x-2)))+C