Final answer to the problem
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-8e^{3x}x+\frac{8}{3}e^{3x}+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}+C_0$
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Step-by-step Solution
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Integrate by parts Integrate by partial fractions Integrate by substitution 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
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1
We can solve the integral $\int\left(3x^2+e^{3x}\right)^3dx$ 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 $3x$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part
$u=3x$
Intermediate steps
2
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
$du=3dx$
Explain this step further
3
Isolate $dx$ in the previous equation
$du=3dx$
Intermediate steps
4
Rewriting $x$ in terms of $u$
$x=\frac{u}{3}$
Explain this step further
Intermediate steps
5
Substituting $u$, $dx$ and $x$ in the integral and simplify
$\int\frac{\left(\frac{1}{3}u^2+e^u\right)^3}{3}du$
Explain this step further
6
Take the constant $\frac{1}{3}$ out of the integral
$\frac{1}{3}\int\left(\frac{1}{3}u^2+e^u\right)^3du$
$\frac{1}{3}\int\left(\frac{1}{3}u^2+e^u\right)^3du$
Intermediate steps
8
Rewrite the integrand $\left(\frac{1}{3}u^2+e^u\right)^3$ in expanded form
$\frac{1}{3}\int\left(\frac{1}{27}u^{6}+\frac{1}{3}u^{4}e^u+u^2e^{2u}+e^{3u}\right)du$
Explain this step further
9
Expand the integral $\int\left(\frac{1}{27}u^{6}+\frac{1}{3}u^{4}e^u+u^2e^{2u}+e^{3u}\right)du$ into $4$ integrals using the sum rule for integrals, to then solve each integral separately
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{3}\int\frac{1}{3}u^{4}e^udu+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
10
We can solve the integral $\int\frac{1}{3}u^{4}e^udu$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
11
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=\frac{1}{3}u^{4}}\\ \displaystyle{du=\frac{4}{3}u^{3}du}\end{matrix}$
Explain this step further
12
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
14
The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
$e^u$
Intermediate steps
15
Now replace the values of $u$, $du$ and $v$ in the last formula
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{3}\left(\frac{1}{3}e^uu^{4}-\frac{4}{3}\int u^{3}e^udu\right)+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
Explain this step further
16
Multiply the single term $\frac{1}{3}$ by each term of the polynomial $\left(\frac{1}{3}e^uu^{4}-\frac{4}{3}\int u^{3}e^udu\right)$
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}\int u^{3}e^udu+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
17
We can solve the integral $\int u^{3}e^udu$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
18
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=u^{3}}\\ \displaystyle{du=3u^{2}du}\end{matrix}$
Explain this step further
19
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
21
The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
$e^u$
Intermediate steps
22
Now replace the values of $u$, $du$ and $v$ in the last formula
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}\left(e^u\cdot u^{3}-3\int u^{2}e^udu\right)+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
Explain this step further
23
Multiply the single term $-\frac{4}{9}$ by each term of the polynomial $\left(e^u\cdot u^{3}-3\int u^{2}e^udu\right)$
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}\int u^{2}e^udu+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
24
We can solve the integral $\int u^{2}e^udu$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
25
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=u^{2}}\\ \displaystyle{du=2udu}\end{matrix}$
Explain this step further
26
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
28
The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
$e^u$
Intermediate steps
29
Now replace the values of $u$, $du$ and $v$ in the last formula
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}\left(e^u\cdot u^{2}-2\int ue^udu\right)+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
Explain this step further
30
Multiply the single term $\frac{4}{3}$ by each term of the polynomial $\left(e^u\cdot u^{2}-2\int ue^udu\right)$
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\int ue^udu+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
31
We can solve the integral $\int ue^udu$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula
$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$
Intermediate steps
32
First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=u}\\ \displaystyle{du=du}\end{matrix}$
Explain this step further
33
Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=e^udu}\\ \displaystyle{\int dv=\int e^udu}\end{matrix}$
35
The integral of the exponential function is given by the following formula $\displaystyle \int a^xdx=\frac{a^x}{\ln(a)}$, where $a > 0$ and $a \neq 1$
$e^u$
36
Now replace the values of $u$, $du$ and $v$ in the last formula
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int e^{3u}du$
37
We can solve the integral $\int e^{3u}du$ 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 $v$), which when substituted makes the integral easier. We see that $3u$ it's a good candidate for substitution. Let's define a variable $v$ and assign it to the choosen part
$v=3u$
Intermediate steps
38
Now, in order to rewrite $du$ in terms of $dv$, we need to find the derivative of $v$. We need to calculate $dv$, we can do that by deriving the equation above
$dv=3du$
Explain this step further
39
Isolate $du$ in the previous equation
$dv=3du$
40
Substituting $v$ and $du$ in the integral and simplify
$\frac{1}{3}\int\frac{1}{27}u^{6}du+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}\int u^2e^{2u}du+\frac{1}{3}\int\frac{e^v}{3}dv$
Intermediate steps
41
The integral $\frac{1}{3}\int\frac{1}{27}u^{6}du$ results in: $\frac{1}{567}u^{7}$
$\frac{1}{567}u^{7}$
Explain this step further
Intermediate steps
42
The integral $\frac{1}{3}\int u^2e^{2u}du$ results in: $\frac{1}{3}u^{3}e^{2u}$
$\frac{1}{3}u^{3}e^{2u}$
Explain this step further
Intermediate steps
43
The integral $\frac{1}{3}\int\frac{e^v}{3}dv$ results in: $\frac{1}{9}e^{9x}$
$\frac{1}{9}e^{9x}$
Explain this step further
44
Gather the results of all integrals
$\frac{1}{567}u^{7}+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}u^{3}e^{2u}+\frac{1}{9}e^{9x}$
Intermediate steps
45
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^uu^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}u^{3}e^{2u}+\frac{1}{9}e^{9x}$
Explain this step further
Intermediate steps
46
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^u\cdot u^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}u^{3}e^{2u}+\frac{1}{9}e^{9x}$
Explain this step further
Intermediate steps
47
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^u\cdot u^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}u^{3}e^{2u}+\frac{1}{9}e^{9x}$
Explain this step further
Intermediate steps
48
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}u^{3}e^{2u}+\frac{1}{9}e^{9x}$
Explain this step further
Intermediate steps
49
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}$
Explain this step further
50
Solve the product $-\frac{8}{3}\left(e^u\cdot u-\int e^udu\right)$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-\frac{8}{3}e^u\cdot u+\frac{8}{3}\int e^udu+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}$
Intermediate steps
51
Replace $u$ with the value that we assigned to it in the beginning: $3x$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-8e^{3x}x+\frac{8}{3}\int e^udu+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}$
Explain this step further
Intermediate steps
52
The integral $\frac{8}{3}\int e^udu$ results in: $\frac{8}{3}e^{3x}$
$\frac{8}{3}e^{3x}$
Explain this step further
53
Gather the results of all integrals
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-8e^{3x}x+\frac{8}{3}e^{3x}+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}$
54
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-8e^{3x}x+\frac{8}{3}e^{3x}+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}+C_0$
Final answer to the problem
$\frac{1}{567}\left(3x\right)^{7}+\frac{1}{9}e^{3x}\left(3x\right)^{4}-\frac{4}{9}e^{3x}\left(3x\right)^{3}+\frac{4}{3}e^{3x}\left(3x\right)^{2}-8e^{3x}x+\frac{8}{3}e^{3x}+\frac{1}{3}\left(3x\right)^{3}e^{6x}+\frac{1}{9}e^{9x}+C_0$