To integrate this function, use u substitution. Make,

 \(\displaystyle \\u=x^2+1 \\du=2xdx\) 

then substitute them into the equation to get

 \(\displaystyle f(x)=\frac{du}{u}\).

The integral of

 \(\displaystyle \frac{du}{u}=ln|u|+C\)

then plug u back into the equation

\(\displaystyle ln|x^2+1|+C\).

The +C is essential because the integral is indefinite.

To integrate this function, use u substitution. Make

 \(\displaystyle \\u=sec(x) \\du=sec(x)tan(x)dx\) 

then substitute them into the equation to get

 \(\displaystyle \int udu\).

The integral of

 \(\displaystyle udu=\frac{u^2}{2}+C\)

then plug u back into the equation

\(\displaystyle \frac{sec^2(x)}{2}+C\).

The +C is essential because the integral is indefinite.

To integrate this function, use u substitution. Make

 \(\displaystyle \\u=1-x^2 \\du=-2xdx\) 

then substitute them into the equation to get

 \(\displaystyle \int \frac{\frac{1}{2}du}{\sqrt{u}}\).

The integral of

 \(\displaystyle \int \frac{du}{\sqrt{u}}=2\sqrt{u}+C\)

so we have

\(\displaystyle \frac{1}{2}\cdot2\sqrt{u}+C=\sqrt{u}+C=\sqrt{1-x^2}+C\). 

The +C is essential because the integral is indefinite.

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 

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