What is the integral of (xe^x)/(sqrt(4+e^x)) ?

The integral of (xe^x)/(sqrt(4+e^x)) is

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integral of (xe^x)/(sqrt(4+e^x))

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

2 x 4 + e^{x} + 8 (\frac{1}{2} ln \frac{1}{2} 4 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{2} 4 + e^{x} - 1 - \frac{1}{2} 4 + e^{x}) + C

Solution steps

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

Apply u-substitution

: \int \frac{ln ( u )}{4 + u} d u

= \int \frac{ln ( u )}{4 + u} d u

Apply Integration By Parts

: 2 ln (u) 4 + u - \int \frac{2 4 + u}{u} d u

= 2 ln (u) 4 + u - \int \frac{2 4 + u}{u} d u

\int \frac{2 4 + u}{u} d u = - 8 (\frac{1}{2} ln \frac{4 + u}{2} + 1 - \frac{1}{2} ln \frac{4 + u}{2} - 1 - \frac{1}{2} 4 + u)

= 2 ln (u) 4 + u - (- 8 (\frac{1}{2} ln \frac{4 + u}{2} + 1 - \frac{1}{2} ln \frac{4 + u}{2} - 1 - \frac{1}{2} 4 + u))

Substitute back

u = e^{x}

= 2 ln (e^{x}) 4 + e^{x} - (- 8 (\frac{1}{2} ln \frac{4 + e ^{x}}{2} + 1 - \frac{1}{2} ln \frac{4 + e ^{x}}{2} - 1 - \frac{1}{2} 4 + e^{x}))

Simplify

2 ln (e^{x}) 4 + e^{x} - (- 8 (\frac{1}{2} ln \frac{4 + e ^{x}}{2} + 1 - \frac{1}{2} ln \frac{4 + e ^{x}}{2} - 1 - \frac{1}{2} 4 + e^{x})) : 2 x 4 + e^{x} + 8 (\frac{1}{2} ln \frac{1}{2} 4 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{2} 4 + e^{x} - 1 - \frac{1}{2} 4 + e^{x})

= 2 x 4 + e^{x} + 8 (\frac{1}{2} ln \frac{1}{2} 4 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{2} 4 + e^{x} - 1 - \frac{1}{2} 4 + e^{x})

Add a constant to the solution

= 2 x 4 + e^{x} + 8 (\frac{1}{2} ln \frac{1}{2} 4 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{2} 4 + e^{x} - 1 - \frac{1}{2} 4 + e^{x}) + C

\int \frac{3 - z}{2 + z} d z

2 e^{x}

\int \frac{4 x}{( x - 3 ) ^{2}} d x

\int \frac{x}{4 - x} d x

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

What is the integral of (xe^x)/(sqrt(4+e^x)) ?
The integral of (xe^x)/(sqrt(4+e^x)) is