Find the integral $\int\frac{6x}{x^3-8}dx$

Step-by-step Solution

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Final answer to the problem

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

Rewrite the expression $\frac{6x}{x^3-8}$ inside the integral in factored form

$\int\frac{6x}{\left(x-2\right)\left(x^2+2x+4\right)}dx$

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

$\int\frac{6x}{\left(x-2\right)\left(x^2+2x+4\right)}dx$

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Learn how to solve integrals by partial fraction expansion problems step by step online. Find the integral int((6x)/(x^3-8))dx. Rewrite the expression \frac{6x}{x^3-8} inside the integral in factored form. Take out the constant 6 from the integral. Rewrite the fraction \frac{x}{\left(x-2\right)\left(x^2+2x+4\right)} in 2 simpler fractions using partial fraction decomposition. Expand the integral \int\left(\frac{1}{6\left(x-2\right)}+\frac{-\frac{1}{6}x+\frac{1}{3}}{x^2+2x+4}\right)dx into 2 integrals using the sum rule for integrals, to then solve each integral separately.

Final answer to the problem

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

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Function Plot

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

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5
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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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