Find the integral $\int\frac{3x^3-x^2+6x-4}{\left(x^2+1\right)\left(x^2+2\right)}dx$

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Final answer to the problem

$-3\arctan\left(x\right)+\frac{3}{2}\ln\left|x^2+1\right|+2\cdot \left(\frac{1}{\sqrt{2}}\right)\arctan\left(\frac{x}{\sqrt{2}}\right)+C_0$
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Step-by-step Solution

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  • Integrate by partial fractions
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Rewrite the fraction $\frac{3x^3-x^2+6x-4}{\left(x^2+1\right)\left(x^2+2\right)}$ in $2$ simpler fractions using partial fraction decomposition

$\frac{3x-3}{x^2+1}+\frac{2}{x^2+2}$

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

$\frac{3x-3}{x^2+1}+\frac{2}{x^2+2}$

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Learn how to solve integrals by partial fraction expansion problems step by step online. Find the integral int((3x^3-x^26x+-4)/((x^2+1)(x^2+2)))dx. Rewrite the fraction \frac{3x^3-x^2+6x-4}{\left(x^2+1\right)\left(x^2+2\right)} in 2 simpler fractions using partial fraction decomposition. Expand the integral \int\left(\frac{3x-3}{x^2+1}+\frac{2}{x^2+2}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int\frac{3x-3}{x^2+1}dx results in: \frac{3}{2}\ln\left(x^2+1\right)-3\arctan\left(x\right). Gather the results of all integrals.

Final answer to the problem

$-3\arctan\left(x\right)+\frac{3}{2}\ln\left|x^2+1\right|+2\cdot \left(\frac{1}{\sqrt{2}}\right)\arctan\left(\frac{x}{\sqrt{2}}\right)+C_0$

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Function Plot

Plotting: $-3\arctan\left(x\right)+\frac{3}{2}\ln\left(x^2+1\right)+2\cdot \left(\frac{1}{\sqrt{2}}\right)\arctan\left(\frac{x}{\sqrt{2}}\right)+C_0$

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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 (3)

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