What is the integral of (\sqrt[5]{2x+1}) ?

The integral of (\sqrt[5]{2x+1}) is 5/12 (2x+1)^{6/5}+C

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integral of (\sqrt[5]{2x+1})

\frac{5}{12} (2 x + 1)^{\frac{6}{5}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 2 x + 1

= \frac{1}{2} \cdot \frac{5}{6} (2 x + 1)^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{6} (2 x + 1)^{\frac{6}{5}} : \frac{5}{12} (2 x + 1)^{\frac{6}{5}}

= \frac{5}{12} (2 x + 1)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{12} (2 x + 1)^{\frac{6}{5}} + C

y^{^{''}} = 9 t^{2}

\int cos (n) π x d x

\int (π - x) d x

\int x + 9 d x

\int e^{- 3 x + 2} d x

What is the integral of (\sqrt[5]{2x+1}) ?
The integral of (\sqrt[5]{2x+1}) is 5/12 (2x+1)^{6/5}+C