# joHn eDwArd's question at Yahoo! Answers regarding an indefinite integral

#### MarkFL

##### Administrator
Staff member
Here is the question:

How to get the integral of (x^3 + 2x + sin x) e^(3x) dx?

please include the complete solution. thanks
I have posted a link there to this topic so the OP can see my work.

#### MarkFL

##### Administrator
Staff member
Hello joHn eDwArd,

We are given to evaluate:

$$\displaystyle \int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx$$

I would write:

$$\displaystyle y=\int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx$$

Now, differentiating with respect to $x$, we obtain the ODE:

$$\displaystyle \frac{dy}{dx}=\left(x^3+2x+\sin(x) \right)e^{3x}$$

We see the homogeneous solution is:

$$\displaystyle y_h(x)=c_1$$

and, using the method of undetermined coefficients, we should expect a particular solution of the form:

$$\displaystyle y_p(x)=\left(Ax^3+Bx^2+Cx+D+E\cos(x)+F\sin(x) \right)e^{3x}$$

Differentiating with respect to $x$ and substituting into the ODE, we obtain after dividing though by $e^{3x}$:

$$\displaystyle 3Ax^3+(3A+3B)x^2+(2B+3C)x+(C+3D)+(3F-E)\sin(x)+(3E+F)\cos(x)=x^3+2x+\sin(x)$$

Equating coefficients, we find:

$$\displaystyle 3A=1\,\therefore\,A=\frac{1}{3}$$

$$\displaystyle 3A+3B=0\,\therefore\,B=-A=-\frac{1}{3}$$

$$\displaystyle 2B+3C=2\,\therefore\,C=\frac{8}{9}$$

$$\displaystyle C+3D=0\,\therefore\,D=-\frac{C}{3}=-\frac{8}{27}$$

$$\displaystyle 3F-E=1$$

$$\displaystyle 3E+F=0\,\therefore\,E=-\frac{1}{10},\,F=\frac{3}{10}$$

And so we have:

$$\displaystyle y_p(x)=\left(\frac{1}{3}x^3-\frac{1}{3}x^2+\frac{8}{9}x-\frac{8}{27}-\frac{1}{10}\cos(x)+\frac{3}{10}\sin(x) \right)e^{3x}$$

$$\displaystyle y_p(x)=\frac{e^{3x}}{270}\left(90x^3-90x^2+240x-80-27\cos(x)+81\sin(x) \right)$$

And so by superposition, we have:

$$\displaystyle y(x)=y_h(x)+y_p(x)$$

Hence, we may conclude:

$$\displaystyle \int\left(x^3+2x+\sin(x) \right)e^{3x}\,dx=\frac{e^{3x}}{270}\left(90x^3-90x^2+240x-80-27\cos(x)+81\sin(x) \right)+C$$

#### Prove It

##### Well-known member
MHB Math Helper
Here is the question:

I have posted a link there to this topic so the OP can see my work.
Here's how I would do the problem:

\displaystyle \displaystyle \begin{align*} \int{ \left[ x^3 + 2x + \sin{(x)} \right] e^{3x}\,dx} = \int{ \left( x^3 + 2x \right) e^{3x} \,dx} + \int{ \sin{(x)}e^{3x}\,dx} \end{align*}

So now we can evaluate each integral by using Integration by Parts.

\displaystyle \begin{align*} \int{ \left( x^3 + 2x \right) e^{3x}\,dx} &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \int{ \frac{1}{3} \left( 3x^2 + 2 \right) e^{3x}\,dx } \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{3} \left[ \frac{1}{3} \left( 3x^2 + 2 \right) e^{3x} - \int{ \frac{1}{3} \left( 6x \right) e^{3x} \,dx } \right] \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{3} \int{ x\,e^{3x}\,dx} \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{ 2}{3} \left[ \frac{1}{3} x\,e^{3x} - \int{ \frac{1}{3} e^{3x}\,dx } \right] \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{9} \int{ e^{3x}\,dx} \\ &= \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{27}e^{3x} \end{align*}

and

\displaystyle \begin{align*} I &= \int{\sin{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \int{ \frac{1}{3}\cos{(x)}e^{3x}\,dx } \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{3} \int{ \cos{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{3} \left[ \frac{1}{3}\cos{(x)}e^{3x} - \int{ -\frac{1}{3}\sin{(x)}e^{3x}\,dx } \right] \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} - \frac{1}{9}\int{\sin{(x)}e^{3x}\,dx} \\ I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} - \frac{1}{9} I \\ \frac{10}{9}I &= \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} \\ I &= \frac{9}{10} \left[ \frac{1}{3}\sin{(x)}e^{3x} - \frac{1}{9}\cos{(x)}e^{3x} \right] \\ I &= \frac{3}{10}\sin{(x)}e^{3x} - \frac{1}{10}\cos{(x)}e^{3x} \end{align*}

Therefore \displaystyle \begin{align*} \int{ \left[ x^3 + 2x + \sin{(x)} \right] e^{3x}\,dx}\end{align*}

\displaystyle \begin{align*} = \frac{1}{3} \left( x^3 + 2x \right) e^{3x} - \frac{1}{9} \left( 3x^2 + 2 \right) e^{3x} + \frac{2}{9}x\,e^{3x} - \frac{2}{27}e^{3x} + \frac{3}{10}\sin{(x)}e^{3x} - \frac{1}{10}\cos{(x)}e^{3x} + C \end{align*}

#### MarkFL

##### Administrator
Staff member
Here's how I would do the problem:...
I truly appreciate you taking the time to present this so nicely! The OP is probably expected to use IBP here, but I got "lazy" and decided to treat it as an ODE instead.

#### Prove It

##### Well-known member
MHB Math Helper
I truly appreciate you taking the time to present this so nicely! The OP is probably expected to use IBP here, but I got "lazy" and decided to treat it as an ODE instead.
Don't worry Mark, your solution is quite beautiful, and you weren't the only lazy one. I could not be arsed taking out common factors and collecting like terms (if there are any) hahaha.