What is the integral of (6x)/(\sqrt[5]{(x^2+7)^2)} ?

The integral of (6x)/(\sqrt[5]{(x^2+7)^2)} is 5(x^2+7)^{3/5}+C

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integral of (6x)/(\sqrt[5]{(x^2+7)^2)}

5 (x^{2} + 7)^{\frac{3}{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

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

Take the constant out:

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

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

Apply the Power Rule

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

Substitute back

u = x^{2} + 7

= 6 \cdot \frac{1}{2} \cdot \frac{5}{3} (x^{2} + 7)^{\frac{3}{5}}

Simplify

6 \cdot \frac{1}{2} \cdot \frac{5}{3} (x^{2} + 7)^{\frac{3}{5}} : 5 (x^{2} + 7)^{\frac{3}{5}}

= 5 (x^{2} + 7)^{\frac{3}{5}}

Add a constant to the solution

= 5 (x^{2} + 7)^{\frac{3}{5}} + C

What is the integral of (6x)/(\sqrt[5]{(x^2+7)^2)} ?
The integral of (6x)/(\sqrt[5]{(x^2+7)^2)} is 5(x^2+7)^{3/5}+C