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Find the integral $\int\left(3x^2+1\right)^6dx$

Step-by-step Solution

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Final Answer

$\frac{729}{13}x^{13}+\frac{1458}{11}x^{11}+135x^{9}+\frac{540}{7}x^{7}+27x^{5}+6x^{3}+x+C_0$
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Step-by-step Solution

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We can expand the expression inside the integral $\left(3x^2+1\right)^6$ using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer $n$. The formula is as follows: $\displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n$. The number of terms resulting from the expansion always equals $n + 1$. The coefficients $\left(\begin{matrix}n\\k\end{matrix}\right)$ are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of $a$ decreases, from $n$ to $0$, while the exponent of $b$ increases, from $0$ to $n$. If one of the binomial terms is negative, the positive and negative signs alternate.

$\int\left(729x^{12}+1458x^{10}+1215x^{8}+540x^{6}+135x^{4}+18x^2+1\right)dx$

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$\int\left(729x^{12}+1458x^{10}+1215x^{8}+540x^{6}+135x^{4}+18x^2+1\right)dx$

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Learn how to solve problems step by step online. Find the integral int((3x^2+1)^6)dx. We can expand the expression inside the integral \left(3x^2+1\right)^6 using Newton's binomial theorem, which is a formula that allow us to find the expanded form of a binomial raised to a positive integer n. The formula is as follows: \displaystyle(a\pm b)^n=\sum_{k=0}^{n}\left(\begin{matrix}n\\k\end{matrix}\right)a^{n-k}b^k=\left(\begin{matrix}n\\0\end{matrix}\right)a^n\pm\left(\begin{matrix}n\\1\end{matrix}\right)a^{n-1}b+\left(\begin{matrix}n\\2\end{matrix}\right)a^{n-2}b^2\pm\dots\pm\left(\begin{matrix}n\\n\end{matrix}\right)b^n. The number of terms resulting from the expansion always equals n + 1. The coefficients \left(\begin{matrix}n\\k\end{matrix}\right) are combinatorial numbers which correspond to the nth row of the Tartaglia triangle (or Pascal's triangle). In the formula, we can observe that the exponent of a decreases, from n to 0, while the exponent of b increases, from 0 to n. If one of the binomial terms is negative, the positive and negative signs alternate.. Expand the integral \int\left(729x^{12}+1458x^{10}+1215x^{8}+540x^{6}+135x^{4}+18x^2+1\right)dx into 7 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int729x^{12}dx results in: \frac{729}{13}x^{13}. The integral \int1458x^{10}dx results in: \frac{1458}{11}x^{11}.

Final Answer

$\frac{729}{13}x^{13}+\frac{1458}{11}x^{11}+135x^{9}+\frac{540}{7}x^{7}+27x^{5}+6x^{3}+x+C_0$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of ((3x^2+1)^6)dx using partial fractionsSolve integral of ((3x^2+1)^6)dx using basic integralsSolve integral of ((3x^2+1)^6)dx using u-substitutionSolve integral of ((3x^2+1)^6)dx using integration by partsSolve integral of ((3x^2+1)^6)dx using trigonometric substitution

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Plotting: $\frac{729}{13}x^{13}+\frac{1458}{11}x^{11}+135x^{9}+\frac{540}{7}x^{7}+27x^{5}+6x^{3}+x+C_0$

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0
a
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g
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n
u
v
w
x
y
z
.
(◻)
+
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×
◻/◻
/
÷
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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