Compute ∫ 2 3 t ( ln ⁡ ( 4 t ) − 1 ) d t.

Since&nbsp;23&nbsp;is constant with respect to&nbsp;t, move&nbsp;23&nbsp;out of the integral.23∫1t(ln(4t)-1)dtLet&nbsp;u2=ln(4t). Then&nbsp;du2=1tdt, so&nbsp;tdu2=dt. Rewrite using&nbsp;u2&nbsp;and&nbsp;du2.23∫1u2-1du2Let&nbsp;u3=u2-1. Then&nbsp;du3=du2. Rewrite using&nbsp;u3&nbsp;and&nbsp;du3.23∫1u3du3The integral of&nbsp;1u3&nbsp;with respect to&nbsp;u3&nbsp;is&nbsp;ln(|u3|).23(ln(|u3|)+C)Simplify.23ln(|u3|)+CSubstitute back in for each integration substitution variable.23ln(|(4t)-1|)+C

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Answered question

2022-05-11

Compute

\int \frac{2}{3 t (\ln (4 t) - 1)} ﻿ d t

Answer & Explanation

necrologo9yh43

Beginner2022-05-12Added 23 answers

Since $\frac{2}{3}$ is constant with respect to $t$ , move $\frac{2}{3}$ out of the integral.

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

Let $u_{2} = \ln (4 t)$ . Then $d u_{2} = \frac{1}{t} d t$ , so $t d u_{2} = d t$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

$\frac{2}{3} \int \frac{1}{u_{2} - 1} d u_{2}$

Let $u_{3} = u_{2} - 1$ . Then $d u_{3} = d u_{2}$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

$\frac{2}{3} \int \frac{1}{u_{3}} d u_{3}$

The integral of $\frac{1}{u_{3}}$ with respect to $u_{3}$ is $\ln (| u_{3} |)$ .

$\frac{2}{3} (\ln (| u_{3} |) + C)$

Simplify.

$\frac{2}{3} \ln (| u_{3} |) + C$

Substitute back in for each integration substitution variable.

$\frac{2}{3} \ln (| (4 t) - 1 |) + C$

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