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Step-by-step Solution

Integrate (2x-1)/((x+3)^2) from 0 to 2

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Answer

$\left(2\ln\left|x+3\right|+2\int_{0}^{2}-3u^{-2}du\right)-\frac{2}{15}$

Step-by-step explanation

Problem to solve:

$\int_0^2\left(\left(\frac{2x-1}{\left(x+3\right)^2}\right)\right)dx$
1

Split the fraction $\frac{2x+-1}{\left(x+3\right)^2}$ in two terms with same denominator

$\int_{0}^{2}\left(\frac{2x}{\left(x+3\right)^2}+\frac{-1}{\left(x+3\right)^2}\right)dx$
2

The integral of a sum of two or more functions is equal to the sum of their integrals

$\int_{0}^{2}\frac{2x}{\left(x+3\right)^2}dx+\int_{0}^{2}\frac{-1}{\left(x+3\right)^2}dx$

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Answer

$\left(2\ln\left|x+3\right|+2\int_{0}^{2}-3u^{-2}du\right)-\frac{2}{15}$
$\int_0^2\left(\left(\frac{2x-1}{\left(x+3\right)^2}\right)\right)dx$

Main topic:

Integral calculus

Used formulas:

6. See formulas

Time to solve it:

~ 2.04 seconds