What is the integral of x\sqrt[5]{64+x^2} ?

The integral of x\sqrt[5]{64+x^2} is 5/12 (64+x^2)^{6/5}+C

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integral of x\sqrt[5]{64+x^2}

\frac{5}{12} (64 + x^{2})^{\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 = 64 + x^{2}

= \frac{1}{2} \cdot \frac{5}{6} (64 + x^{2})^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{6} (64 + x^{2})^{\frac{6}{5}} : \frac{5}{12} (64 + x^{2})^{\frac{6}{5}}

= \frac{5}{12} (64 + x^{2})^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{12} (64 + x^{2})^{\frac{6}{5}} + C

f (x) = \frac{6}{x}

(10, x), (8, 10)

y = \frac{6 x ^{2} + 8 x + 6}{x}

x + 2, - 1, 2, 5

\frac{d ^{6}}{d x ^{6}} (x^{- 3})

What is the integral of x\sqrt[5]{64+x^2} ?
The integral of x\sqrt[5]{64+x^2} is 5/12 (64+x^2)^{6/5}+C