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)}

\int \frac{( z + 1 )}{3 z ^{2} + 2 z + 2} d z

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

Solution steps

\int \frac{z + 1}{3 z ^{2} + 2 z + 2} d z

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{2 3 v} d v

= \int \frac{1}{2 3 v} d v

Take the constant out:

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

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

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