What is the integral of x\sqrt[8]{64+x^2} ?

The integral of x\sqrt[8]{64+x^2} is 4/9 (x^2+64)^{9/8}+C

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integral of x\sqrt[8]{64+x^2}

\frac{4}{9} (x^{2} + 64)^{\frac{9}{8}} + 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{1}{2} \cdot \frac{8}{9} u^{\frac{9}{8}}

Substitute back

u = 64 + x^{2}

= \frac{1}{2} \cdot \frac{8}{9} (64 + x^{2})^{\frac{9}{8}}

Simplify

\frac{1}{2} \cdot \frac{8}{9} (64 + x^{2})^{\frac{9}{8}} : \frac{4}{9} (x^{2} + 64)^{\frac{9}{8}}

= \frac{4}{9} (x^{2} + 64)^{\frac{9}{8}}

Add a constant to the solution

= \frac{4}{9} (x^{2} + 64)^{\frac{9}{8}} + C

What is the integral of x\sqrt[8]{64+x^2} ?
The integral of x\sqrt[8]{64+x^2} is 4/9 (x^2+64)^{9/8}+C