The integral is found by recognizing the following integral:

\(\displaystyle \int \frac{dx}{x^2+1}=\arctan (x)+C\)

To solve the integral, make the following substitution:

\(\displaystyle u=x+2\), \(\displaystyle du=dx\)

The derivative was found using the following rule:

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

The integral then becomes

\(\displaystyle \int \frac{du}{u^2+1}=\arctan (u)+1\)

Finish by replacing u with the original term containing x.

Substitution can be used to make this problem easier. Let u = sin(x). Then du = cos(x)dx. This allows us to rewrite and evaluate the original integral in terms of u as:

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

The final answer should be written in terms of the original variables, so substitute sin(x) for u. 

\(\displaystyle \frac{u^2}{2}+c=\frac{sin^2(x)}{2}+c\)

Note that we could have also chosen cos(x) as u, but the above substitution avoids introducing negatives. 

First, notice that \(\displaystyle e^{2x}=(e^{x})^2\). Use the substitution \(\displaystyle u=e^x, du=e^xdx\) to rewrite the integral as

\(\displaystyle \int{\frac{1}{1+e^{2x}}dx}=\int{\frac{1}{1+u^{2}}du}\).

Next, recall that \(\displaystyle \int\frac{1}{1+x^2}dx=arctan(x)+c\), so 

\(\displaystyle \int{\frac{1}{1+u^{2}}du}=arctan(u)+c\). 

Lastly, substitute \(\displaystyle e^x\) in place of \(\displaystyle u\) to write the answer in terms of the original variables. 

\(\displaystyle arctan(u)+c=arctan(e^x)+c\)

To solve the indefinite integral, make a simple substitution:

\(\displaystyle u=\cos(x), du=-\sin(x)dx\)

The integral then becomes:

\(\displaystyle \int -u^4du\)

After integrating, we get

\(\displaystyle -\frac{u^5}{5}+C\)

The following rule was used for the integration:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

Finally, replace the u with the original term containing x. 

We know that the derivative of  \(\displaystyle ln(x)\) is  \(\displaystyle \frac{1}x{}\).

Doing a substitution and setting

\(\displaystyle u = \ln x\) and \(\displaystyle du = \frac{dx}{x}\) allows us to rewrite the integral as 

\(\displaystyle \int \frac{du}{u^{2}}\) which can be rewritten as \(\displaystyle \int u^{-2}du\).

Integrating this gets you \(\displaystyle \frac{-1}{u}\) plus a constant (which is stated in the original question that you can assume that we already have one). Substituting \(\displaystyle x\) back in gives us the final answer, which is 

\(\displaystyle \frac{-1}{ln(x)} + C\).

We know that the derivative of \(\displaystyle \arctan(x)\) is \(\displaystyle \frac{1}{1 + x^2}\).

So substituting

\(\displaystyle u = \arctan(x)\) allows us to have 

\(\displaystyle du= \frac{1}{1 + x^2}\).

This allows us to rewrite the integral as 

\(\displaystyle \int u du\) which, when integrated, gives us 

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

Substiting x back in gives us the answer, 

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

To evaluate the integral, we must recognize that what we were given looks very similar to the following integral:

\(\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a})+C\)

To make our integral look like the one above, we must perform the following substiution:

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

Now, rewrite our integral:

\(\displaystyle \frac{1}{3}\int \frac{du}{\sqrt{4-u^2}}\)

It looks like the one above, so we can integrate now:

\(\displaystyle \frac{1}{3}\int \frac{du}{\sqrt{4-u^2}} = \frac{1}{3}\arcsin(\frac{u}{2})+C\)

Finally, replace u with our original term:

\(\displaystyle \frac{1}{3}\arcsin(\frac{3}{2}x)+C\)

To integrate, we must break the integral into three integrals:

\(\displaystyle \int x^4dx +\int \frac{dx}{x}+\int \frac{dx}{4x^2+`1}\)

The first integral is equal to 

\(\displaystyle \frac{x^5}{5}+C\)

and was found using the following rule:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

The second integral is equal to

\(\displaystyle \ln\left | x\right |+C\)

and was found using the following rule:

\(\displaystyle \int \frac{dx}{x}=\ln \left | x\right |+C\)

The final integral is found by performing the following substitution:

\(\displaystyle u=2x, du=2dx\)

Now, rewrite and integrate:

\(\displaystyle \frac{1}{2}\int \frac{du}{u^2+1}=\frac{\arctan(u)}{2}+C\)

The integral was found using the following rule:

\(\displaystyle \int \frac{dx}{x^2+1}=\arctan(x)+C\)

Finally, rewrite the integral in terms of \(\displaystyle x\) by replacing \(\displaystyle u\) with the original term, and add all three integrals together to get a final answer of 

\(\displaystyle \frac{x^5}{5}+\ln\left |x \right |+\frac{\arctan(2x)}{2}+C\)

To integrate, we must make the following substitution:

\(\displaystyle u=\sin(x), du=\cos(x)\)

The derivative was found using the following rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)\)

Now, rewrite the integral:

\(\displaystyle \int (1-u^2)(u^5)du\)

Notice that we changed

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

Next, distribute and integrate:

\(\displaystyle \int (u^5-u^7)du=\frac{u^6}{6}-\frac{u^8}{8}+C\)

The integral was found using the following rule:

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\)

Finally, replace \(\displaystyle u\) with our original \(\displaystyle x\) term:

\(\displaystyle -\frac{\sin^8(x)}{8}+\frac{\sin^6(x)}{6}+C\)

To solve the integral, we must make the following substitution:

\(\displaystyle u=\sin(x), du=\cos(x)\)

Now, rewrite the integral and integrate:

\(\displaystyle \int u^6 du=\frac{u^7}{7} + C\)

The integral was found using the following rule:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

Finally, replace \(\displaystyle u\) with our original \(\displaystyle x\) term:

\(\displaystyle \frac{\sin^7(x)}{7}+C\)