To calculate the integral, we need to use integration by parts. The definition for integration by parts is

\(\displaystyle \int udv=uv-\int vdu\)

It is important here to select the correct u and dv terms from our orginal integral. We eventually want of the terms to "go away" when we take its derivate. We notice here that out of our two functions in our integral, \(\displaystyle x\) and \(\displaystyle e^{3x}\), the derivate of x is 1, making is very simple to integrate eventually. Therefore, \(\displaystyle x\) will be our \(\displaystyle u\) term, and \(\displaystyle e^{3x}dx\) will be our dv term. Note that the dv term is not just dx, but the function attached to it as well. If \(\displaystyle e^{3x}\) was our \(\displaystyle u\) term, then \(\displaystyle xdx\) would be our dv term.

\(\displaystyle u=x\)                      \(\displaystyle dv=e^{3x}dx\)

Now calculate the terms \(\displaystyle du\) and \(\displaystyle v\) needed to proceed with the integration by parts equation.

\(\displaystyle du=dx\)                 \(\displaystyle v=\int e^{3x}dx=\frac{1}{3}e^{3x}\) 

(you can to set integration constant c=0)

Now that we have the terms that we need, we can plug in these terms into the integration by parts formula above.

\(\displaystyle (x)\left(\frac{1}{3}e^{3x}\right)\) - \(\displaystyle \int \frac{1}{3}e^{3x}dx\)

Note that although we still need to integrate one more time, this new integral only consists of one function which is simple to integrate, as opposed to the two functions we had before. Also note that the x term from the initial integral "went away", thus making the resulting integral easy to calculate.

Simplifying this term now becomes

\(\displaystyle \frac{1}{3}xe^{3x}+\frac{1}{9}e^{3x}+c\).

Although the integral looks difficult, it can be majorly simplified. Remember this crucial trig identity.

\(\displaystyle 1-cos^2(u)=sin^2(u)\)

Using this identity, the integral can now be simplified to

\(\displaystyle \int\sqrt{2sin^2(\sqrt2x)}dx=\sqrt2 \int sin(x)dx\), which is very simple to integrate.

\(\displaystyle =\sqrt2 \int sin(\sqrt2 x)dx\)

\(\displaystyle =-cos(\sqrt2x)+c\)

In this problem we can try to get all the terms with \(\displaystyle y\) on one side and all the terms with \(\displaystyle x\) on the other.

\(\displaystyle ydy=e^xdx\)

Now we can integrate both sides using the definition of \(\displaystyle e\) and the power rule.

\(\displaystyle \int x^n=\frac{x^{n+1}}{n+1}\) and \(\displaystyle \int e^x=e^x\)

\(\displaystyle \int ydy=\int e^xdx\)

\(\displaystyle \frac{y^2}{2}=e^x +C,C=0\)

\(\displaystyle y^2=2e^x\)

\(\displaystyle y=\pm\sqrt{2e^x}\)

Move all expressions with \(\displaystyle y\) to one side and all \(\displaystyle x\) to the other.

\(\displaystyle y{dy}=\frac{dx}{x+2}\)

Now integrate both sides using the power rule and the definition of natural log.

The power rule for integrals states, 

\(\displaystyle \int x^n =\frac{x^{n+1}}{n+1}\) and the definition of natural log is,

\(\displaystyle \int \frac{1}{x}=ln(x)\).

Applying these rules we are able to solve the problem.

\(\displaystyle \frac{y^2}{2}=ln(x+2)\)

\(\displaystyle y=\pm\sqrt{2ln(x+2)}\)

We need to use the following identity:

\(\displaystyle sin^2(x)=\frac{1-cos(2x)}{2}\) 

Our integral now becomes

\(\displaystyle \frac{1}{2} \int \1-cos(4x)dx\).

Notice inside the cosine becomes 4 because we already had a 2 in the original expression.

This can be split into two integrals

\(\displaystyle \frac{1}{2} \int 1dx +\frac{1}{2}\int \cos(4x)dx\).

Which becomes 

\(\displaystyle \frac{1}{2}\left(x-\frac{sin({4x)}}4\right)+C\).

We can find the indefinite integral of \(\displaystyle f'(x)=9x^{2}+6x-12\)  using the Power Rule for Integrals, which states that 

\(\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C\) for all \(\displaystyle n\neq-1\) and with the arbitrary constant of integration \(\displaystyle C\).

Applying this rule to 

\(\displaystyle \int f'(x)dx=\int (9x^{2}+6x-12)dx=3x^{3}+3x^{2}-12x+C\). 

We can find the indefinite integral of \(\displaystyle f'(x)=12x^{2}-2x+7\)  using the Power Rule for Integrals, which states that 

\(\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C\) for all \(\displaystyle n\neq-1\) and with the arbitrary constant of integration \(\displaystyle C\).

Applying this rule to 

\(\displaystyle \int f'(x)dx=\int (12x^{2}-2x+7)dx=4x^{3}-x^{2}+7x+C\). 

Since \(\displaystyle \int x^{2}\cdot \sin (x) dx\) doesn't resemble any basic integral form, we can use Integration by Parts, which follows this pattern.

\(\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du\)

Where \(\displaystyle \int u \cdot dv\) is representing \(\displaystyle \int x^{2}\cdot \sin (x) dx\).

In general, we choose "u" to be the part of the integral that will eventually turn to zero if we differentiate it repeatedly, and "dv" to be the rest of the integral. In this case,

\(\displaystyle u = x^{2}\) , and \(\displaystyle dv = \sin (x) dx\).

Now we differentiate "u" to find "du", and integrate "dv" to find "v". Doing so gives,

\(\displaystyle du = \frac{\mathrm{d} }{\mathrm{d} x}(x^{2}) \cdot dx = 2x \cdot dx\), and \(\displaystyle v = \int \sin (x) dx = - \cos (x)\).

Now we plug the pieces into the Integration by Parts pattern.

\(\displaystyle {\color{Red} \int u \cdot dv} ={\color{Green} u} \cdot {\color{Blue} v} - \int {\color{Blue} v} \cdot {\color{Green} du}\)

\(\displaystyle {\color{Red} \int x^{2}\cdot \sin (x) dx} = {\color{Green} x^{2} }\cdot {\color{Blue} (-\cos x)} - \int {\color{Blue} (-\cos x)} \cdot {\color{Green} 2x \cdot dx}\)

Simplifying a little gives the following.

\(\displaystyle \int x^{2}\cdot \sin (x) dx = -x^{2} \cdot \cos x +2 \int x \cdot (\cos x) dx\).

Unfortunately, this remaining integral, \(\displaystyle \int x \cdot (\cos x) dx\), is still not in a form that we can integrate. However, if we apply Integration by Parts to this second integral, we will get something that we can integrate. Using the same reasoning as before, choose "u" and "dv".

\(\displaystyle u = x\)                                                   \(\displaystyle dv = (\cos x) dx\)

Then find "du" and "v" by the same method as before.

\(\displaystyle du = \frac{\mathrm{d} }{\mathrm{d} x}(x) dx= (1)dx\)                 \(\displaystyle v = \int (\cos x) dx = \sin x\)

\(\displaystyle du = dx\)                                             \(\displaystyle v= \sin x\)

Now, put all the pieces together for the second iteration of Integration by Parts.

\(\displaystyle {\color{Red} \int u \cdot dv} ={\color{Green} u} \cdot {\color{Blue} v} - \int {\color{Blue} v} \cdot {\color{Green} du}\)

\(\displaystyle {\color{Red} \int x \cdot (\cos x) dx} = {\color{Green} x} \cdot {\color{Blue} \sin x}- \int {\color{Blue} (\sin x)}{\color{Green} dx}\)

Now we have an integral that resembles a basic integral form. Let's integrate it.

\(\displaystyle \int x \cdot (\cos x)dx= x \cdot \sin x - {\color{Red} \int (\sin x)dx}\)

\(\displaystyle \int x \cdot (\cos x)dx= x \cdot \sin x - [{\color{Red} (-\cos x) + C}]\)

Let's simplify this result before substituting it back into the first Integration by parts equation.

\(\displaystyle \int x \cdot (\cos x)dx= x \cdot \sin x + \cos x + C\)

Now substitute this into the first Integration by parts result.

\(\displaystyle \int x^{2}\cdot \sin (x) dx = -x^{2} \cdot \cos x +2{\color{Red} \int x \cdot (\cos x) dx}\)

\(\displaystyle \int x^{2}\cdot \sin (x) dx = -x^{2} \cdot \cos x +2[{\color{Red} x \cdot \sin x + \cos x + C}]\)

Now we simplify.

\(\displaystyle \int x^{2}\cdot \sin (x) dx = -x^{2} \cdot \cos x +2 x \cdot \sin x + 2 \cdot \cos x + C\)

Thus, the final answer is

\(\displaystyle -x^{2} \cdot \cos x +2 x \cdot \sin x + 2 \cdot \cos x + C\)

Neither \(\displaystyle e^{x}\) nor \(\displaystyle \sin (2x)\) will reduce to zero no matter how many times we differentiate them. So we must use Integration by Parts twice and solve for the integral algebraicly to find the answer. The process starts by choosing "u" and "dv". In this case, both \(\displaystyle e^{x}\) and \(\displaystyle \sin (2x)\) can be both differentiated and integrated without issue, so it doesn't matter which is "u" and which is "dv". In this example, we will choose the following.

\(\displaystyle u = \sin (2x)\)                                            \(\displaystyle dv = e^{x} dx\)

Then we find "du" by differentiating "u", and we find "v" by integrating "dv".

\(\displaystyle du = \frac{\mathrm{d} }{\mathrm{d} x} (\sin (2x))dx\)                           \(\displaystyle v = \int e^{x}dx\)

\(\displaystyle du = 2 \cdot \cos (2x) dx\)                              \(\displaystyle v = e^{x}\)

Now we assemble these pieces into the Integration by Parts pattern.

\(\displaystyle {\color{Red} \int u \cdot dv} = {\color{Green} u} \cdot {\color{Blue} v} - \int {\color{Blue} v} \cdot {\color{Green} du}\)

\(\displaystyle {\color{Red} \int \sin (2x)\cdot e^{x}dx} = {\color{Green} \sin (2x)} \cdot {\color{Blue} e^{x}} - \int {\color{Blue} e^{x}} \cdot {\color{Green} 2 \cdot \cos (2x) dx}\)

Let's simplify a little and pull the constant, 2, out of the new integral.

\(\displaystyle \int \sin (2x) \cdot e^{x} dx= \sin (2x) \cdot e^{x} -2 \int e^{x} \cdot \cos (2x) dx\)

Now we have to use Integration by parts on the new integral, \(\displaystyle \int e^{x} \cdot \cos (2x) dx\), with "u" and "dv" chosen similarly to the first Integration by Parts.

\(\displaystyle u = \cos (2x)\)                                     \(\displaystyle dv = e^{x} dx\)

Now differentiate "u" and integrate "dv".

\(\displaystyle du = -2 \cdot \sin (2x) dx\)                     \(\displaystyle v = e^{x}\)

Now apply Integration by parts to \(\displaystyle \int e^{x} \cdot \cos (2x) dx\).

\(\displaystyle {\color{Red} \int u \cdot dv} = {\color{Green} u} \cdot {\color{Blue} v} - \int {\color{Blue} v} \cdot {\color{Green} du}\)

\(\displaystyle {\color{Red} \int \cos (2x) \cdot e^{x} dx} = {\color{Green} \cos (2x)} \cdot {\color{Blue} e^{x}}- \int {\color{Blue} e^{x}} \cdot{\color{Green} (-2 \cdot \sin (2x)) dx}\)

Lets simplify the new integral.

\(\displaystyle \int e^{x} \cdot \cos (2x) dx = e^{x} \cdot \cos (2x) +2 \int e^{x} \cdot \sin (2x) dx\)

Substituting this back into the first Integration by Parts equation, we get

\(\displaystyle \int e^{x} \cdot \sin (2x) dx= \sin (2x) \cdot e^{x} -2{\color{Red} [e^{x}\cdot \cos (2x) + 2 \int e^{x} \cdot \sin (2x) dx]}\)

Now simplify by mulitplying the -2 throught the substituted expression.

\(\displaystyle \int e^{x} \cdot \sin (2x) dx= \sin (2x) \cdot e^{x} -2e^{x}\cdot \cos (2x) - 4 \int e^{x} \cdot \sin (2x) dx\)

Now notice that the original integral we are evaluating is now present on both sides of the equation. Solve for it by adding \(\displaystyle 4 \int e^{x} \cdot \sin (2x) dx\) to both sides, thus canceling the right sides's, then combine them on the left side.

\(\displaystyle \int e^{x} \cdot \sin (2x) dx {\color{Red} +4\int e^{x} \cdot \sin (2x)dx}= \sin (2x) \cdot e^{x} -2e^{x}\cdot \cos (2x)\)

\(\displaystyle 5\int e^{x} \cdot \sin (2x) dx = \sin (2x) \cdot e^{x} -2e^{x}\cdot \cos (2x)\)

Now isolate the integral by dividing both sides by 5. This cancels the coefficient of 5, and leaves the integral that we are evaluating isolated on the left side of the equation.

\(\displaystyle \int e^{x} \cdot \sin (2x) dx =\frac{e^{x} \cdot \sin (2x) -2e^{x}\cdot \cos (2x)}{5}+C\)

We added the constant of integration into our final answer because this is an indefinite integral. This gives the final answer.

\(\displaystyle \frac{e^{x} \cdot \sin (2x) -2e^{x}\cdot \cos (2x)}{5}+C\)

In order to integrate, we must use integration by parts, which follows the formula

\(\displaystyle \int udv=uv-\int vdu\)

We must set our \(\displaystyle u\) to be a function easy to differentiate:

\(\displaystyle u=3x, du=3dx\)

This makes the integration step very easy later on!

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

The rest of our given integrand is \(\displaystyle dv\):

\(\displaystyle dv=\cos(15x), v=\frac{\sin(15x)}{15}\)

The integral was found using the following rule:

\(\displaystyle \int \cos(x)dx=\sin(x)+C\)

(We substitute

\(\displaystyle g=15x, dg=15dx\) 

into the integral and integrate according to the above rule. Finally we replace \(\displaystyle g\) with our \(\displaystyle x\) term from above.)

Now, write out the integration by parts:

\(\displaystyle \frac{x\sin(15x)}{5}-\int \frac{\sin(15x)}{5}dx\)

Finally, integrate:

\(\displaystyle \int \frac{\sin(15x)}{5}dx=-\frac{\cos(15x)}{5}+C\)

The integration was performed using the following rule:

\(\displaystyle \int \sin(x)dx=-\cos(x)+C\)

Our final answer is henceforth

\(\displaystyle \frac{x\sin(15x)}{5}+\frac{\cos(15x)}{5}+C\)