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Find the integral $\int\frac{x-2}{\left(2x+1\right)\left(x+3\right)}dx$

Step-by-step Solution

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Final answer to the problem

$-\frac{1}{2}\ln\left|2x+1\right|+\ln\left|x+3\right|+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
  • Integrate by substitution
  • Integrate by parts
  • Integrate using tabular integration
  • Integrate by trigonometric substitution
  • Weierstrass Substitution
  • Integrate using trigonometric identities
  • Integrate using basic integrals
  • Product of Binomials with Common Term
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Rewrite the fraction $\frac{x-2}{\left(2x+1\right)\left(x+3\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{x-2}{\left(2x+1\right)\left(x+3\right)}=\frac{A}{2x+1}+\frac{B}{x+3}$

Learn how to solve integrals by partial fraction expansion problems step by step online.

$\frac{x-2}{\left(2x+1\right)\left(x+3\right)}=\frac{A}{2x+1}+\frac{B}{x+3}$

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Learn how to solve integrals by partial fraction expansion problems step by step online. Find the integral int((x-2)/((2x+1)(x+3)))dx. Rewrite the fraction \frac{x-2}{\left(2x+1\right)\left(x+3\right)} in 2 simpler fractions using partial fraction decomposition. Find the values for the unknown coefficients: A, B. The first step is to multiply both sides of the equation from the previous step by \left(2x+1\right)\left(x+3\right). Multiplying polynomials. Simplifying.

Final answer to the problem

$-\frac{1}{2}\ln\left|2x+1\right|+\ln\left|x+3\right|+C_0$

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Function Plot

Plotting: $-\frac{1}{2}\ln\left|2x+1\right|+\ln\left|x+3\right|+C_0$

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Integrals by Partial Fraction Expansion

The partial fraction decomposition or partial fraction expansion of a rational function is the operation that consists in expressing the fraction as a sum of a polynomial (possibly zero) and one or several fractions with a simpler denominator.

Used Formulas

See formulas (4)

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