Evaluate ∫ 3 t 5 2 t − 1 d t.

Question

Evaluate   ∫            3      t              5              2        t        −        1                  ﻿    d  t.

Kosyging1j7u · Accepted Answer

This integral could not be completed using integration by parts. Mathway will use another method.

Since $\frac{3}{5}$ is constant with respect to $t$ , move $\frac{3}{5}$ out of the integral.

$\frac{3}{5} \int \frac{t}{\sqrt{2 t - 1}} d t$

Let $u = 2 t - 1$ . Then $d u = 2 d t$ , so $\frac{1}{2} d u = d t$ . Rewrite using $u$ and $d$ $u$ .

$\frac{3}{5} \int \frac{\frac{u}{2} + \frac{1}{2}}{\sqrt{u}} \cdot \frac{1}{2} d u$

Simplify.

$\frac{3}{5} \int \frac{u + 1}{4 \sqrt{u}} d u$

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

$\frac{3}{5} (\frac{1}{4} \int \frac{u + 1}{\sqrt{u}} d u)$

Simplify the expression.

$\frac{3}{20} \int (u + 1) u^{- \frac{1}{2}} d u$

Expand $(u + 1) u^{- \frac{1}{2}}$ .

$\frac{3}{20} \int u^{\frac{1}{2}} + u^{- \frac{1}{2}} d u$

Split the single integral into multiple integrals.

$\frac{3}{20} (\int u^{\frac{1}{2}} d u + \int u^{- \frac{1}{2}} d u)$

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

$\frac{3}{20} (\frac{2}{3} u^{\frac{3}{2}} + C + \int u^{- \frac{1}{2}} d u)$

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

$\frac{3}{20} (\frac{2}{3} u^{\frac{3}{2}} + C + 2 u^{\frac{1}{2}} + C)$

Simplify.

$\frac{3}{20} (\frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}}) + C$

Replace all occurrences of $u$ with $2 t - 1$ .

$\frac{3}{20} (\frac{2}{3} {(2 t - 1)}^{\frac{3}{2}} + 2 {(2 t - 1)}^{\frac{1}{2}}) + C$

Simplify.

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

Evaluate <mrow> 3 t </mrow> <mrow> 5 <msqrt> 2