What is the integral of (e^x)/(\sqrt[5]{e^x+1)} ?

The integral of (e^x)/(\sqrt[5]{e^x+1)} is 5/4 (e^x+1)^{4/5}+C

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integral of (e^x)/(\sqrt[5]{e^x+1)}

\frac{5}{4} (e^{x} + 1)^{\frac{4}{5}} + C

Solution steps

Apply u-substitution

Apply the Power Rule

: \frac{5}{4} u^{\frac{4}{5}}

= \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = e^{x} + 1

= \frac{5}{4} (e^{x} + 1)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{4} (e^{x} + 1)^{\frac{4}{5}} + C

y = x^{2} cos^{4} (x)

\int (tan (2 x)) d x

\frac{d}{d x} (\frac{1}{8} (- x^{2} + 8 ln (x)))

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

\int_{0}^{7} k (7 - x)

What is the integral of (e^x)/(\sqrt[5]{e^x+1)} ?
The integral of (e^x)/(\sqrt[5]{e^x+1)} is 5/4 (e^x+1)^{4/5}+C