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

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

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

\frac{3}{4} (z^{2} + 2 z + 2)^{\frac{2}{3}} + 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{3}{2} v^{\frac{2}{3}}

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{3}{2} ((z + 1)^{2} + 1)^{\frac{2}{3}} : \frac{3}{4} (z^{2} + 2 z + 2)^{\frac{2}{3}}

= \frac{3}{4} (z^{2} + 2 z + 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{4} (z^{2} + 2 z + 2)^{\frac{2}{3}} + C

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