What is the integral of \sqrt[3]{(x^2+1)^4}x ?

The integral of \sqrt[3]{(x^2+1)^4}x is 3/14 (x^2+1)^{7/3}+C

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integral of \sqrt[3]{(x^2+1)^4}x

\frac{3}{14} (x^{2} + 1)^{\frac{7}{3}} + C

Solution steps

Apply u-substitution

: \int \frac{( u + 1 ) ^{\frac{4}{3}}}{2} d u

= \int \frac{( u + 1 ) ^{\frac{4}{3}}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int (u + 1)^{\frac{4}{3}} d u

Apply u-substitution

: \int v^{\frac{4}{3}} d v

= \frac{1}{2} \cdot \int v^{\frac{4}{3}} d v

Apply the Power Rule

: \frac{3}{7} v^{\frac{7}{3}}

= \frac{1}{2} \cdot \frac{3}{7} v^{\frac{7}{3}}

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{3}{7} (x^{2} + 1)^{\frac{7}{3}} : \frac{3}{14} (x^{2} + 1)^{\frac{7}{3}}

= \frac{3}{14} (x^{2} + 1)^{\frac{7}{3}}

Add a constant to the solution

= \frac{3}{14} (x^{2} + 1)^{\frac{7}{3}} + C

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What is the integral of \sqrt[3]{(x^2+1)^4}x ?
The integral of \sqrt[3]{(x^2+1)^4}x is 3/14 (x^2+1)^{7/3}+C