What is the integral of (\sqrt[5]{4x^2+5})/3 x ?

The integral of (\sqrt[5]{4x^2+5})/3 x is 5/144 (4x^2+5)^{6/5}+C

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integral of (\sqrt[5]{4x^2+5})/3 x

\frac{5}{144} (4 x^{2} + 5)^{\frac{6}{5}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

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} v^{\frac{6}{5}}

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{5}{6} v^{\frac{6}{5}}

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{5}{6} (4 x^{2} + 5)^{\frac{6}{5}}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{5}{6} (4 x^{2} + 5)^{\frac{6}{5}} : \frac{5}{144} (4 x^{2} + 5)^{\frac{6}{5}}

= \frac{5}{144} (4 x^{2} + 5)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{144} (4 x^{2} + 5)^{\frac{6}{5}} + C

n = 1 \sum \infty n^{(- \frac{2}{3})}

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

x - 2

x \to - 5 lim (x^{2})

What is the integral of (\sqrt[5]{4x^2+5})/3 x ?
The integral of (\sqrt[5]{4x^2+5})/3 x is 5/144 (4x^2+5)^{6/5}+C