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

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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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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. Expand the integral \int\left(\frac{-1}{2x+1}+\frac{1}{x+3}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int\frac{-1}{2x+1}dx results in: -\frac{1}{2}\ln\left(2x+1\right). The integral \int\frac{1}{x+3}dx results in: \ln\left(x+3\right).

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
D□x
∫
∫◻
|◻|
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

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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.

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