Kiersten Hodge
2022-05-13
Answered

Compute $\int \frac{4t}{5\sqrt[3]{2t+3}}dt$.

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Darion Sexton

Answered 2022-05-14
Author has **15** answers

Since $\frac{4}{5}$ is constant with respect to $t$, move $\frac{4}{5}$ out of the integral.

$\frac{4}{5}\int \frac{t}{\sqrt[3]{2t+3}}dt$

Let $u=2t+3$. Then $du=2dt$, so $\frac{1}{2}du=dt$. Rewrite using $u$ and $d$$u$.

$\frac{4}{5}\int \frac{\frac{u}{2}-\frac{3}{2}}{\sqrt[3]{u}}\cdot \frac{1}{2}du$

Simplify.

$\frac{4}{5}\int \frac{u-3}{4\sqrt[3]{u}}du$

Since $\frac{1}{4}$ is constant with respect to $u$, move $\frac{1}{4}$ out of the integral.

$\frac{4}{5}(\frac{1}{4}\int \frac{u-3}{\sqrt[3]{u}}du)$

Simplify the expression.

$\frac{1}{5}\int (u-3){u}^{-\frac{1}{3}}du$

Expand $(u-3){u}^{-\frac{1}{3}}$.

$\frac{1}{5}\int {u}^{\frac{2}{3}}-3{u}^{-\frac{1}{3}}du$

Split the single integral into multiple integrals.

$\frac{1}{5}(\int {u}^{\frac{2}{3}}du+\int -3{u}^{-\frac{1}{3}}du)$

By the Power Rule, the integral of $u}^{\frac{2}{3}$ with respect to $u$ is $\frac{3}{5}{u}^{\frac{5}{3}}$.

$\frac{1}{5}(\frac{3}{5}{u}^{\frac{5}{3}}+C+\int -3{u}^{-\frac{1}{3}}du)$

Since $-3$ is constant with respect to $u$, move $-3$ out of the integral.

$\frac{1}{5}(\frac{3}{5}{u}^{\frac{5}{3}}+C-3\int {u}^{-\frac{1}{3}}du)$

By the Power Rule, the integral of $u}^{-\frac{1}{3}$ with respect to $u$ is $\frac{3}{2}{u}^{\frac{2}{3}}$.

$\frac{1}{5}(\frac{3}{5}{u}^{\frac{5}{3}}+C-3(\frac{3}{2}{u}^{\frac{2}{3}}+C))$

Simplify.

$\frac{1}{5}(\frac{3{u}^{\frac{5}{3}}}{5}-\frac{9{u}^{\frac{2}{3}}}{2})+C$

Replace all occurrences of $u$ with $2t+3$.

$\frac{1}{5}(\frac{3{(2t+3)}^{\frac{5}{3}}}{5}-\frac{9{(2t+3)}^{\frac{2}{3}}}{2})+C$

Simplify.

$\frac{3{(2t+3)}^{{\displaystyle \frac{2}{3}}}(4t-9)}{50}+C$

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