"Evaluate the indefinite integral. ∫z21+z33dz" solved and explained

Falak Kinney

Answered question

2021-10-27

Evaluate the indefinite integral.
$\int \frac{z^{2}}{\sqrt[3]{1 + z^{3}}} d z$

Answer & Explanation

Arnold Odonnell

Skilled2021-10-28Added 109 answers

Step 1
Consider the provided integral,
$\int \frac{z^{2}}{\sqrt[3]{1 + z^{3}}} d z$
Evaluate the indefinite integral.
Apply the substitution method,
Let,
$u = 1 + z^{3} \Rightarrow \frac{d u}{d z} = 3 z^{2}$
$\Rightarrow \frac{1}{3} d u = z^{2} d z$
Step 2
Therefore, the integrals becomes,
$\int \frac{z^{2}}{\sqrt[3]{1 + z^{3}}} d z = \int \frac{1}{3 \sqrt[3]{u}} d u$
$= \frac{1}{3} \cdot \int \frac{1}{3 \sqrt[3]{u}} d u$
$= \frac{1}{3} \cdot \int \frac{1}{u^{\frac{1}{3}}} d u$
$= \frac{1}{3} \cdot \frac{u^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1} + C$
Step 3
Substitute back $u = 1 + z^{3}$ .
$\int \frac{z^{2}}{\sqrt[3]{1 + z^{2}}} d z = \frac{1}{3} \cdot \frac{{(1 + z^{3})}^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1}$
$= \frac{1}{2} {(1 + z^{3})}^{\frac{2}{3}} + C$
Hence.

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