What is the integral of (e^xsqrt(e^x-1))/(e^x+8) ?

The integral of (e^xsqrt(e^x-1))/(e^x+8) is 6(-arctan(1/3 sqrt(e^x-1))+1/3 sqrt(e^x-1))+C

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integral of (e^xsqrt(e^x-1))/(e^x+8)

\int \frac{e ^{x} e ^{x} - 1}{e ^{x} + 8} d x

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

Solution steps

\int \frac{e ^{x} e ^{x} - 1}{e ^{x} + 8} d x

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 9} d u

= \int \frac{2 u ^{2}}{u ^{2} + 9} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} + 9} d u

\frac{u ^{2}}{u ^{2} + 9} = - \frac{9}{u ^{2} + 9} + 1

= 2 \cdot \int - \frac{9}{u ^{2} + 9} + 1 d u

Apply Integral Substitution

= 2 \cdot \int 3 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 3 (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

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

\int 1 d v = v

= 2 \cdot 3 (- arctan (v) + v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (e^xsqrt(e^x-1))/(e^x+8) ?
The integral of (e^xsqrt(e^x-1))/(e^x+8) is 6(-arctan(1/3 sqrt(e^x-1))+1/3 sqrt(e^x-1))+C