Compute ∫ 2 3 t t + 1 3 d t.

Integrate by parts using the formula&nbsp;∫udv=uv-∫vdu, where&nbsp;u=23t&nbsp;and&nbsp;dv=t+13.23t(34(t+1)43)-∫34(t+1)4323dtSimplify.t(t+1)432-∫(t+1)432dtSince&nbsp;12&nbsp;is constant with respect to&nbsp;t, move&nbsp;12&nbsp;out of the integral.t(t+1)432-(12∫(t+1)43dt)Let&nbsp;u=t+1. Then&nbsp;du=dt. Rewrite using&nbsp;u&nbsp;and&nbsp;du.t(t+1)432-12∫u43duBy the Power Rule, the integral of&nbsp;u43&nbsp;with respect to&nbsp;u&nbsp;is&nbsp;37u73.t(t+1)432-12(37u73+C)Simplify.7t(t+1)43-3u7314+CReplace all occurrences of&nbsp;u&nbsp;with&nbsp;t+1.7t(t+1)43-3(t+1)7314+CReorder terms.114(7t(t+1)43-3(t+1)73)+C

Step by step solution to "Compute ∫ 2 3 t t +..."

Alisa Durham

Answered question

2022-05-11

Compute

\int \frac{2}{3} ﻿ t \sqrt[3]{t + 1} d t

Answer & Explanation

Timothy Mcclure

Beginner2022-05-12Added 15 answers

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \frac{2}{3} t$ and $d v = \sqrt[3]{t + 1}$ .

$\frac{2}{3} t (\frac{3}{4} {(t + 1)}^{\frac{4}{3}}) - \int \frac{3}{4} {(t + 1)}^{\frac{4}{3}} \frac{2}{3} d t$

Simplify.

$\frac{t {(t + 1)}^{\frac{4}{3}}}{2} - \int \frac{{(t + 1)}^{\frac{4}{3}}}{2} d t$

Since $\frac{1}{2}$ is constant with respect to $t$ , move $\frac{1}{2}$ out of the integral.

$\frac{t {(t + 1)}^{\frac{4}{3}}}{2} - (\frac{1}{2} \int {(t + 1)}^{\frac{4}{3}} d t)$

Let $u = t + 1$ . Then $d u = d t$ . Rewrite using $u$ and $d$ $u$ .

$\frac{t {(t + 1)}^{\frac{4}{3}}}{2} - \frac{1}{2} \int u^{\frac{4}{3}} d u$

By the Power Rule, the integral of $u^{\frac{4}{3}}$ with respect to $u$ is $\frac{3}{7} u^{\frac{7}{3}}$ .

$\frac{t {(t + 1)}^{\frac{4}{3}}}{2} - \frac{1}{2} (\frac{3}{7} u^{\frac{7}{3}} + C)$

Simplify.

$\frac{7 t {(t + 1)}^{\frac{4}{3}} - 3 u^{\frac{7}{3}}}{14} + C$

Replace all occurrences of $u$ with $t + 1$ .

$\frac{7 t {(t + 1)}^{\frac{4}{3}} - 3 {(t + 1)}^{\frac{7}{3}}}{14} + C$

Reorder terms.

$\frac{1}{14} (7 t {(t + 1)}^{\frac{4}{3}} - 3 {(t + 1)}^{\frac{7}{3}}) + C$

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