prsategazd

2021-12-28

Evaluate the following integrals. Include absolute values only when needed.
${\int }_{0}^{3}\frac{2x-1}{x+1}dx$

### Answer & Explanation

Thomas Lynn

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 $x+1=u⇒dx=du$
Also, when $x=0⇒u=1$ & when $x=3⇒u=4$
So, the integral becomes:
${\int }_{1}^{4}\frac{2\left(u-1\right)-1}{u}du$
$⇒{\int }_{1}^{4}\frac{2u-3}{u}du$
$⇒{\int }_{1}^{4}\frac{2u}{u}-\frac{3}{u}du$
$⇒{\int }_{1}^{4}2-\frac{3}{u}du$
$=2u-3\mathrm{ln}\left(u\right){\mid }_{1}^{4}$
$=2\left(4\right)-3\mathrm{ln}\left(4\right)-2\left(1\right)+3\mathrm{ln}\left(1\right)$
$⇒8-3\mathrm{ln}\left(4\right)-2+3\left(0\right)$
$⇒6-3\mathrm{ln}\left({2}^{2}\right)$
$⇒6-6\mathrm{ln}\left(2\right)$
Hence, ${\int }_{0}^{3}\frac{2x-1}{x+1}dx=6-6\mathrm{ln}\left(2\right)$

zurilomk4

$\int \frac{2x-1}{x+1}dx$
Substitution $u=x+1⇒\frac{du}{dx}=1⇒dx=du$:
$=\int \frac{2u-3}{u}du$
We use the distributive property:
$=\int \left(2-\frac{3}{u}\right)du$
Lets

karton

$\begin{array}{}{\int }_{0}^{3}\frac{2x-1}{x+1}dx\\ \int \frac{2x-1}{x+1}dx\\ \int \frac{2t-3}{t}dt\\ \int \frac{2t}{t}-\frac{3}{t}dt\\ \int 2-\frac{3}{t}dt\\ \int 2dt-\int \frac{3}{t}dt\\ 2t-3\mathrm{ln}\left(|t|\right)\\ 2\left(x+1\right)-3\mathrm{ln}\left(|x+1|\right)\\ 2x=2-3\mathrm{ln}\left(|x+1|\right)\\ 2x+2-3\mathrm{ln}\left(|x+1|\right)\\ \left(2x+2-3\mathrm{ln}\left(|x+1|\right)\right){|}_{0}^{3}\\ 2×3+2-3\mathrm{ln}\left(|3+1|\right)-\left(2×0+2-3\mathrm{ln}\left(|0+1|\right)\right)\end{array}$
Simplify
$6-6\mathrm{ln}\left(2\right)$

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