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Integrate the function $\left(6x+1\right)\left(2x^3+x\right)^2$ from 0 to $4$

Step-by-step Solution

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Final answer to the problem

$223578.8190476$
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Step-by-step Solution

How should I solve this problem?

  • 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
  • FOIL Method
  • Load more...
Can't find a method? Tell us so we can add it.
1

Rewrite the expression $\left(6x+1\right)\left(2x^3+x\right)^2$ inside the integral in factored form

$\int_{0}^{4}\left(6x+1\right)x^2\left(2x^2+1\right)^2dx$

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$\int_{0}^{4}\left(6x+1\right)x^2\left(2x^2+1\right)^2dx$

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Learn how to solve problems step by step online. Integrate the function (6x+1)(2x^3+x)^2 from 0 to 4. Rewrite the expression \left(6x+1\right)\left(2x^3+x\right)^2 inside the integral in factored form. Expand the expression \left(2x^2+1\right)^2 using the square of a binomial: (a+b)^2=a^2+2ab+b^2. Multiplying polynomials 6x+1 and 4x^{4}+4x^2+1. Solve the product 4\left(6x+1\right)x^{4}.

Final answer to the problem

$223578.8190476$

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Function Plot

Plotting: $\left(6x+1\right)\left(2x^3+x\right)^2$

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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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