prsategazd
2021-12-28
Answered

Evaluate the following integrals. Include absolute values only when needed.

${\int}_{0}^{3}\frac{2x-1}{x+1}dx$

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Thomas Lynn

Answered 2021-12-29
Author has **28** answers

Step 1:To determine

Evaluate:

${\int}_{0}^{3}\frac{2x-1}{x+1}dx$

Step 2:Calculation

Consider the given integral:

${\int}_{0}^{3}\frac{2x-1}{x+1}dx$

Let$x+1=u\Rightarrow dx=du$

Also, when$x=0\Rightarrow u=1$ & when $x=3\Rightarrow u=4$

So, the integral becomes:

${\int}_{1}^{4}\frac{2(u-1)-1}{u}du$

$\Rightarrow {\int}_{1}^{4}\frac{2u-3}{u}du$

$\Rightarrow {\int}_{1}^{4}\frac{2u}{u}-\frac{3}{u}du$

$\Rightarrow {\int}_{1}^{4}2-\frac{3}{u}du$

$=2u-3\mathrm{ln}\left(u\right){\mid}_{1}^{4}$

$=2\left(4\right)-3\mathrm{ln}\left(4\right)-2\left(1\right)+3\mathrm{ln}\left(1\right)$

$\Rightarrow 8-3\mathrm{ln}\left(4\right)-2+3\left(0\right)$

$\Rightarrow 6-3\mathrm{ln}\left({2}^{2}\right)$

$\Rightarrow 6-6\mathrm{ln}\left(2\right)$

Hence,${\int}_{0}^{3}\frac{2x-1}{x+1}dx=6-6\mathrm{ln}\left(2\right)$

Evaluate:

Step 2:Calculation

Consider the given integral:

Let

Also, when

So, the integral becomes:

Hence,

zurilomk4

Answered 2021-12-30
Author has **35** answers

Substitution

We use the distributive property:

Lets

karton

Answered 2022-01-04
Author has **368** answers

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