To solve the following integral, we must make a substitution to create the following general form:

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

We make the following subsitution:

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

The derivative was found using the following rule:

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

The integral now looks like this:

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

Notice that it is in the same form as the integral we want.

Now use the form from above to integrate:

\(\displaystyle \frac{1}{2}\sin^{-1}(u)+C\)

To finish the problem, replace u with 2x:

\(\displaystyle \frac{1}{2}\sin^{-1}(2x)+C\).

We can simplify

\(\displaystyle \small \int (\ln x)^2 dx\) 

by first doing a substitution, with \(\displaystyle \small u=\ln x\), which gives us \(\displaystyle \small e^u=x\), which means that \(\displaystyle \small e^udu=dx\). So the integral becomes

\(\displaystyle \small \small \int (\ln x)^2 dx=\int u^2e^u du\)

The integral\(\displaystyle \small \small \int u^2 e^u du\) can be solved using two integration by parts, which will give us 

\(\displaystyle \small \small \small \small \int u^2 e^u du=e^u(u^2-2u+2)+C\)

So now we just plg in \(\displaystyle \small \ln x\) into \(\displaystyle \small u\) and \(\displaystyle \small x\) into \(\displaystyle \small e^u\) to get

\(\displaystyle \small \small \small \small \int (\ln x)^2 dx=\int u^2e^u du=e^u(u^2-2u+2)+C=x((\ln x )^2-2\ln x+2)+C\)

Pull out the constant out in front of the integral.

\(\displaystyle \int_{2}^{6}\frac{2}{3xln^3(x)} \:dx=\frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx\)

Use U-substitution to solve.

\(\displaystyle u=ln(x)\)

\(\displaystyle du=\frac{1}{x}\:dx\)

\(\displaystyle \frac{2}{3}\int_{2}^{6}\frac{1}{xln^3(x)} \:dx= \frac{2}{3}\int u^{-3}\:du= \frac{2}{3}\left[ \frac{u^{-2}}{-2}\right]=-\frac{1}{3}\left[\frac{1}{ln^2(u)}\right]_{2}^{6}\)

\(\displaystyle -\frac{1}{3}\left[\frac{1}{ln^2(6)}-\frac{1}{ln^2(2)}\right]=\frac{1}{3ln^2(2)}-\frac{1}{3ln^2(6)}\)

The integral can be solved with a clever substitution:

\(\displaystyle u=\sqrt{2x+3}, du=(2x+3)^{-\frac{1}{2}}dx\)

The derivative was found using the following rules:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\), \(\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\)

Then, when you rewrite the integral in terms of u, you find that you get:

\(\displaystyle \int du = u+C\)

The integration was performed using the following rule:

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

Finally, replace the u with our original term.

\(\displaystyle \sqrt{2x+3}+C\)

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, 

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