Compute ∫ 6 5 t 2 ln ⁡ ( 1 t ) d t.

Integrate by parts using the formula&nbsp;∫udv=uv-∫vdu, where&nbsp;u=ln(1t)&nbsp;and&nbsp;dv=65t2.ln(1t)(25t3)-∫25t3(-1t)dtSimplify.ln(1t)(2t3)5-∫-2t25dtSince&nbsp;-1&nbsp;is constant with respect to&nbsp;t, move&nbsp;-1&nbsp;out of the integral.ln(1t)(2t3)5--∫2t25dtSimplify.ln(1t)(2t3)5+∫2t25dtSince&nbsp;25&nbsp;is constant with respect to&nbsp;t, move&nbsp;25&nbsp;out of the integral.ln(1t)(2t3)5+25∫t2dtBy the Power Rule, the integral of&nbsp;t2&nbsp;with respect to&nbsp;t&nbsp;is&nbsp;13t3.ln(1t)(2t3)5+25(13t3+C)Simplify the answer.25ln(1t)t3+215t3+C

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Answered question

2022-05-12

Compute

\int \frac{6}{5} ﻿ t^{2} \ln (\frac{1}{t}) d t

Answer & Explanation

Makhi Lyons

Beginner2022-05-13Added 15 answers

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (\frac{1}{t})$ and $d v = \frac{6}{5} t^{2}$ .

$\ln (\frac{1}{t}) (\frac{2}{5} t^{3}) - \int \frac{2}{5} t^{3} (- \frac{1}{t}) d t$

Simplify.

$\frac{\ln (\frac{1}{t}) (2 t^{3})}{5} - \int - \frac{2 t^{2}}{5} d t$

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$\frac{\ln (\frac{1}{t}) (2 t^{3})}{5} - - \int \frac{2 t^{2}}{5} d t$

Simplify.

$\frac{\ln (\frac{1}{t}) (2 t^{3})}{5} + \int \frac{2 t^{2}}{5} d t$

Since $\frac{2}{5}$ is constant with respect to $t$ , move $\frac{2}{5}$ out of the integral.

$\frac{\ln (\frac{1}{t}) (2 t^{3})}{5} + \frac{2}{5} \int t^{2} d t$

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

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

Simplify the answer.

$\frac{2}{5} \ln (\frac{1}{t}) t^{3} + \frac{2}{15} t^{3} + C$

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