What is the integral of (xe^x)/(sqrt(49+e^x)) ?

The integral of (xe^x)/(sqrt(49+e^x)) is

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integral of (xe^x)/(sqrt(49+e^x))

\int \frac{x e ^{x}}{49 + e ^{x}} d x

2 x 49 + e^{x} + 28 (\frac{1}{2} ln \frac{1}{7} 49 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{7} 49 + e^{x} - 1 - \frac{1}{7} 49 + e^{x}) + C

Solution steps

\int \frac{x e ^{x}}{49 + e ^{x}} d x

Apply u-substitution

= \int \frac{ln ( u )}{49 + u} d u

Apply Integration By Parts

= 2 ln (u) 49 + u - \int \frac{2 49 + u}{u} d u

\int \frac{2 49 + u}{u} d u = - 28 (\frac{1}{2} ln \frac{49 + u}{7} + 1 - \frac{1}{2} ln \frac{49 + u}{7} - 1 - \frac{1}{7} 49 + u)

= 2 ln (u) 49 + u - (- 28 (\frac{1}{2} ln \frac{49 + u}{7} + 1 - \frac{1}{2} ln \frac{49 + u}{7} - 1 - \frac{1}{7} 49 + u))

Substitute back

u = e^{x}

= 2 ln (e^{x}) 49 + e^{x} - (- 28 (\frac{1}{2} ln \frac{49 + e ^{x}}{7} + 1 - \frac{1}{2} ln \frac{49 + e ^{x}}{7} - 1 - \frac{1}{7} 49 + e^{x}))

Simplify

2 ln (e^{x}) 49 + e^{x} - (- 28 (\frac{1}{2} ln \frac{49 + e ^{x}}{7} + 1 - \frac{1}{2} ln \frac{49 + e ^{x}}{7} - 1 - \frac{1}{7} 49 + e^{x})) : 2 x 49 + e^{x} + 28 (\frac{1}{2} ln \frac{1}{7} 49 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{7} 49 + e^{x} - 1 - \frac{1}{7} 49 + e^{x})

= 2 x 49 + e^{x} + 28 (\frac{1}{2} ln \frac{1}{7} 49 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{7} 49 + e^{x} - 1 - \frac{1}{7} 49 + e^{x})

Add a constant to the solution

= 2 x 49 + e^{x} + 28 (\frac{1}{2} ln \frac{1}{7} 49 + e^{x} + 1 - \frac{1}{2} ln \frac{1}{7} 49 + e^{x} - 1 - \frac{1}{7} 49 + e^{x}) + C

What is the integral of (xe^x)/(sqrt(49+e^x)) ?
The integral of (xe^x)/(sqrt(49+e^x)) is