What is the integral of (xe^x)/(sqrt(64+e^x)) ?

The integral of (xe^x)/(sqrt(64+e^x)) is

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integral of (xe^x)/(sqrt(64+e^x))

\int \frac{x e ^{x}}{64 + e ^{x}} d x

2 x 64 + e^{x} + 32 (\frac{1}{2} ln \frac{1}{8} 64 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{8} 64 + e^{x} - 1 - \frac{1}{8} 64 + e^{x}) + C

Solution steps

\int \frac{x e ^{x}}{64 + e ^{x}} d x

Apply u-substitution

= \int \frac{ln ( u )}{64 + u} d u

Apply Integration By Parts

= 2 ln (u) 64 + u - \int \frac{2 64 + u}{u} d u

\int \frac{2 64 + u}{u} d u = - 32 (\frac{1}{2} ln \frac{64 + u}{8} + 1 - \frac{1}{2} ln \frac{64 + u}{8} - 1 - \frac{1}{8} 64 + u)

= 2 ln (u) 64 + u - (- 32 (\frac{1}{2} ln \frac{64 + u}{8} + 1 - \frac{1}{2} ln \frac{64 + u}{8} - 1 - \frac{1}{8} 64 + u))

Substitute back

u = e^{x}

= 2 ln (e^{x}) 64 + e^{x} - (- 32 (\frac{1}{2} ln \frac{64 + e ^{x}}{8} + 1 - \frac{1}{2} ln \frac{64 + e ^{x}}{8} - 1 - \frac{1}{8} 64 + e^{x}))

Simplify

2 ln (e^{x}) 64 + e^{x} - (- 32 (\frac{1}{2} ln \frac{64 + e ^{x}}{8} + 1 - \frac{1}{2} ln \frac{64 + e ^{x}}{8} - 1 - \frac{1}{8} 64 + e^{x})) : 2 x 64 + e^{x} + 32 (\frac{1}{2} ln \frac{1}{8} 64 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{8} 64 + e^{x} - 1 - \frac{1}{8} 64 + e^{x})

= 2 x 64 + e^{x} + 32 (\frac{1}{2} ln \frac{1}{8} 64 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{8} 64 + e^{x} - 1 - \frac{1}{8} 64 + e^{x})

Add a constant to the solution

= 2 x 64 + e^{x} + 32 (\frac{1}{2} ln \frac{1}{8} 64 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{8} 64 + e^{x} - 1 - \frac{1}{8} 64 + e^{x}) + C

What is the integral of (xe^x)/(sqrt(64+e^x)) ?
The integral of (xe^x)/(sqrt(64+e^x)) is