Find the integral $\int\frac{24x}{\sqrt[3]{3x^2-4}}dx$

Step-by-step Solution

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Final answer to the problem

$\frac{18\sqrt[3]{\left(x^2-\frac{4}{3}\right)^{2}}}{\sqrt[3]{3}}+C_0$
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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

Take out the constant $24$ from the integral

$24\int\frac{x}{\sqrt[3]{3x^2-4}}dx$

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$24\int\frac{x}{\sqrt[3]{3x^2-4}}dx$

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Unlock the first 3 steps of this solution

Learn how to solve problems step by step online. Find the integral int((24x)/((3x^2-4)^(1/3)))dx. Take out the constant 24 from the integral. First, factor the terms inside the radical by 3 for an easier handling. Taking the constant out of the radical. We can solve the integral 24\int\frac{x}{\sqrt[3]{3}\sqrt[3]{x^2-\frac{4}{3}}}dx by applying integration method of trigonometric substitution using the substitution.

Final answer to the problem

$\frac{18\sqrt[3]{\left(x^2-\frac{4}{3}\right)^{2}}}{\sqrt[3]{3}}+C_0$

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Function Plot

Plotting: $\frac{18\sqrt[3]{\left(x^2-\frac{4}{3}\right)^{2}}}{\sqrt[3]{3}}+C_0$

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1
2
3
4
5
6
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

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