Integral of (y-2)/(y+3)

\int\frac{y-2}{y+3}dy

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Answer

$3+y-5\ln\left|3+y\right|+C_0$

Step by step solution

Problem

$\int\frac{y-2}{y+3}dy$
1

Split the fraction $\frac{y+-2}{3+y}$ in two terms with same denominator

$\int\left(\frac{-2}{3+y}+\frac{y}{3+y}\right)dy$
2

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

$\int\frac{-2}{3+y}dy+\int\frac{y}{3+y}dy$
3

Apply the formula: $\int\frac{n}{b+x}dx$$=n\ln\left|b+x\right|$, where $b=3$, $x=y$ and $n=-2$

$\int\frac{y}{3+y}dy-2\ln\left|3+y\right|$
4

Solve the integral $\int\frac{y}{3+y}dy$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=3+y \\ du=dy\end{matrix}$
5

Rewriting $y$ in terms of $u$

$y=u-3$
6

Substituting $u$, $dy$ and $y$ in the integral

$\int\frac{u-3}{u}du-2\ln\left|3+y\right|$
7

Split the fraction $\frac{u+-3}{u}$ in two terms with same denominator

$\int\left(\frac{-3}{u}+\frac{u}{u}\right)du-2\ln\left|3+y\right|$
8

Simplifying the fraction by $u$

$\int\left(\frac{-3}{u}+1\right)du-2\ln\left|3+y\right|$
9

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

$-2\ln\left|3+y\right|+\int\frac{-3}{u}du+\int1du$
10

The integral of a constant is equal to the constant times the integral's variable

$-2\ln\left|3+y\right|+\int\frac{-3}{u}du+u$
11

Substitute $u$ back for it's value, $3+y$

$-2\ln\left|3+y\right|+\int\frac{-3}{u}du+3+y$
12

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$-2\ln\left|3+y\right|-3\ln\left|u\right|+3+y$
13

Substitute $u$ back for it's value, $3+y$

$-2\ln\left|3+y\right|-3\ln\left|3+y\right|+3+y$
14

Adding $-3\ln\left|3+y\right|$ and $-2\ln\left|3+y\right|$

$3+y-5\ln\left|3+y\right|$
15

Add the constant of integration

$3+y-5\ln\left|3+y\right|+C_0$

Answer

$3+y-5\ln\left|3+y\right|+C_0$

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.32 seconds

Views:

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