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:

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

To evaluate the integral, we first must rewrite it as the following:

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

Now, perform the following subsitution:

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

Next, rewrite the integral and integrate:

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

The integral was performed using the following rule:

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

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

\(\displaystyle \sec(x)+C\)

To evaluate the integral, we must first make the following substitution:

\(\displaystyle u=\sqrt{5}x, du=\sqrt{5}dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{1}{\sqrt{5}}\int \frac{du}{u^2+1}=\frac{1}{\sqrt{5}}\arctan(u)+C\)

The integral was performed using the following rule:

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

Finally, replace u with the x term:

\(\displaystyle \frac{1}{\sqrt{5}}{\arctan(\sqrt{5}x)}+C\)

To integrate, we must first make the following substitution:

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

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

\(\displaystyle \frac{1}{3}\int \frac{du}{u^2+4}=\frac{1}{6}\arctan(\frac{u}{2})+C\)

The integration was performed using the following rule:

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

Finally, replace u with our original term:

\(\displaystyle \frac{1}{6}\arctan(\frac{3x}{2})+C\)

To evaluate the integral, we first must make the following substitution:

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

Now, rewrite the integral, and integrate:

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

We used the following rule for integration:

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

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

\(\displaystyle -\frac{\cos^4(x)}{4}+C\)

To evaluate the integral, first we must make the following substitution:

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

The derivative was found using the following rule:

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

Now, rewrite the integral and integrate:

\(\displaystyle 2\int \frac{du}{u}=2\ln\left | u\right |+C\)

The integral was performed using the following rule:

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

Finally, replace \(\displaystyle u\) with the \(\displaystyle x\) containing term:

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

Note that we removed the absolute value sign because the output of a square root is always positive.