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# Integrate $\int\sqrt[3]{6-2x}dx$

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sin
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asin
acos
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sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

##  Final answer to the problem

$-\frac{3}{8}\sqrt[3]{\left(6-2x\right)^{4}}+C_0$
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##  Step-by-step Solution 

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We can solve the integral $\int\sqrt[3]{6-2x}dx$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $6-2x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=6-2x$

Learn how to solve problems step by step online.

$u=6-2x$

Learn how to solve problems step by step online. Integrate int((6-2x)^1/3)dx. We can solve the integral \int\sqrt[3]{6-2x}dx by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it u), which when substituted makes the integral easier. We see that 6-2x it's a good candidate for substitution. Let's define a variable u and assign it to the choosen part. Now, in order to rewrite dx in terms of du, we need to find the derivative of u. We need to calculate du, we can do that by deriving the equation above. Isolate dx in the previous equation. Substituting u and dx in the integral and simplify.

##  Final answer to the problem

$-\frac{3}{8}\sqrt[3]{\left(6-2x\right)^{4}}+C_0$

##  Explore different ways to solve this problem

SnapXam A2

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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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