What is the integral of (x^2)/(\sqrt[3]{(x^3+1)^5)} ?

The integral of (x^2)/(\sqrt[3]{(x^3+1)^5)} is -1/(2(x^3+1)^{2/3)}+C

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integral of (x^2)/(\sqrt[3]{(x^3+1)^5)}

- \frac{1}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}} + C

Solution steps

Apply u-substitution

: \int \frac{1}{3 u ^{\frac{5}{3}}} d u

= \int \frac{1}{3 u ^{\frac{5}{3}}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u ^{\frac{5}{3}}} d u

Apply the Power Rule

: - \frac{3}{2 u ^{\frac{2}{3}}}

= \frac{1}{3} (- \frac{3}{2 u ^{\frac{2}{3}}})

Substitute back

u = x^{3} + 1

= \frac{1}{3} (- \frac{3}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}})

Simplify

\frac{1}{3} (- \frac{3}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}}) : - \frac{1}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}}

= - \frac{1}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}}

Add a constant to the solution

= - \frac{1}{2 ( x ^{3} + 1 ) ^{\frac{2}{3}}} + C

f (x) = 3 e^{3 x}

\int \frac{sin ( 2 x )}{3 + cos ( 2 x )} d x

1 + 6 t

\frac{d}{d x} (\frac{x ^{2} + 16}{x})

What is the integral of (x^2)/(\sqrt[3]{(x^3+1)^5)} ?
The integral of (x^2)/(\sqrt[3]{(x^3+1)^5)} is -1/(2(x^3+1)^{2/3)}+C