To solve this problem, we have to remember properties of natural logs. Pull the 2 out in front of the integral and solve. Don't forget to add "+C" at the end!

\(\displaystyle \\\int \frac{2}{x}dx=\\ \\ 2\int \frac{1}{x}dx=2*ln(x)\\ \\= 2ln(x)+C\)

To solve this integral, use the power rule. Applying it to this problem gives us the following for the first term:

\(\displaystyle \frac{x^{1+1}}{2*(1+1)}=\frac{x^{2}}{4}\)

And the following for the second term:

\(\displaystyle \frac{5x^{0+1}}{0+1}=\frac{5x}{1}=5x\)

We can combine these terms and add our "C" to get the final answer:

\(\displaystyle \frac{1}{4}x^{2}+5x+C\)

To integrate, we must make the following subsitution:

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

Rearranging to get du in terms of x we get the following.

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

Now, rewrite the integral in terms of u, and integrate:

\(\displaystyle \frac{2}{3}\int {du}=\frac{2}{3}u=\frac{2}{3}\sqrt{3x+1}+C\)

which was found using the following rule:

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

Now, replace u with our original term:

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

Notice that the absolute value went away, because the square root is always positive.

To find this integral, use the method of integration by parts:

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

Let \(\displaystyle u=sin(x),du=cos(x)dx,dv=cos(x)dx,v=sin(x)\) such that

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

There is now an integral on each side; however, it is the same integral. Move the integral on the right to the left:

\(\displaystyle 2\int sin(x)cos(x)dx=sin(x)sin(x)\)

Now recall that we're asked to find the value of \(\displaystyle \int sin(x)cos(x)\), so to find that, we need only divide each side by two:

\(\displaystyle \int sin(x)cos(x)dx=\frac{sin^2(x)}{2}\)

Ta-da.

Begin by rewriting the equation in terms of exponents:

\(\displaystyle \int (x^3+2x^1+(x+2)^\frac{1}{2})dx\)

Afterwards, add one to each exponent and divide by the resultant value for each term to do the integral; be sure to add a constant of integration:

For this particular problem we will use the Power Rule on each term.

Power Rule: \(\displaystyle x^n \rightarrow nx^{n-1}\).

Appying this rule we find the following function. 

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

\(\displaystyle \frac{x^4}{4}+x^2+\frac{2(x+2)^\frac{3}{2}}{3}+C\)

To integrate, we must perform the following substitution:

\(\displaystyle u=x^3, du=3x^2dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{2}{3}\int \cos(u)du = \frac{2}{3}\sin(u)+C\)

The integration was performed using the following rule:

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

Finally, replace u with the original term we designated at the start:

\(\displaystyle \frac{2}{3}\sin(x^3)+C\)

To integrate, we must perform the following substiution:

\(\displaystyle u=4x, du=4dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{1}{4}\int \frac{du}{u^2+1}=\frac{1}{4}\arctan(u)+C\)

The integral was performed using the following rule:

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

Finally, replace u with our original, x term:

\(\displaystyle \frac{1}{4}\arctan(4x)+C\)

To perform the integral, we must use the following substitution:

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

Now, rewrite the integral and integrate:

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

The integration was performed using the following rule:

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

Now, replace u with the original, x containing term:

\(\displaystyle -\cos^2(x)+C\)

To integrate, we must perform the following subsitution:

\(\displaystyle u=\tan(x), du=\sec^2(x)dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \int e^udu=e^u+C\)

We used the following rule to integrate:

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

Finally, replace u with our original term:

\(\displaystyle e^{\tan(x)}+C\)

To integrate, we must first make the following subsitution:

\(\displaystyle u=x^2, du=2xdx\)

Now, rewrite the integral and integrate:

\(\displaystyle 5\int \cos(u)du=5\sin(u)+C\)

We used the following rule to integrate:

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

Finally, replace u with the original term:

\(\displaystyle 5\sin(x^2)+C\)