Since doesn't resemble any basic integral form, we can use Integration by Parts, which follows this pattern.

Where is representing .

In general, we choose "u" to be the part of the integral that will eventually turn to zero if we differentiate it repeatedly, and "dv" to be the rest of the integral. In this case,

, and .

Now we differentiate "u" to find "du", and integrate "dv" to find "v". Doing so gives,

, and .

Now we plug the pieces into the Integration by Parts pattern.

Simplifying a little gives the following.

.

Unfortunately, this remaining integral, , is still not in a form that we can integrate. However, if we apply Integration by Parts to this second integral, we will get something that we can integrate. Using the same reasoning as before, choose "u" and "dv".

Then find "du" and "v" by the same method as before.

Now, put all the pieces together for the second iteration of Integration by Parts.

Now we have an integral that resembles a basic integral form. Let's integrate it.

Let's simplify this result before substituting it back into the first Integration by parts equation.

Now substitute this into the first Integration by parts result.

Now we simplify.

Thus, the final answer is

Neither nor will reduce to zero no matter how many times we differentiate them. So we must use Integration by Parts twice and solve for the integral algebraicly to find the answer. The process starts by choosing "u" and "dv". In this case, both and can be both differentiated and integrated without issue, so it doesn't matter which is "u" and which is "dv". In this example, we will choose the following.

Then we find "du" by differentiating "u", and we find "v" by integrating "dv".

Now we assemble these pieces into the Integration by Parts pattern.

Let's simplify a little and pull the constant, 2, out of the new integral.

Now we have to use Integration by parts on the new integral, , with "u" and "dv" chosen similarly to the first Integration by Parts.

Now differentiate "u" and integrate "dv".

Now apply Integration by parts to .

Lets simplify the new integral.

Substituting this back into the first Integration by Parts equation, we get

Now simplify by mulitplying the -2 throught the substituted expression.

Now notice that the original integral we are evaluating is now present on both sides of the equation. Solve for it by adding to both sides, thus canceling the right sides's, then combine them on the left side.

Now isolate the integral by dividing both sides by 5. This cancels the coefficient of 5, and leaves the integral that we are evaluating isolated on the left side of the equation.

We added the constant of integration into our final answer because this is an indefinite integral. This gives the final answer.

In order to integrate, we must use integration by parts, which follows the formula

We must set our  to be a function easy to differentiate:

This makes the integration step very easy later on!

The derivative was found using the following rule:

The rest of our given integrand is :

The integral was found using the following rule:

(We substitute

into the integral and integrate according to the above rule. Finally we replace  with our  term from above.)

Now, write out the integration by parts:

Finally, integrate:

The integration was performed using the following rule:

Our final answer is henceforth

To evaluate the integral, we must sum two integrals:

The first integral is equal to 

and was found using the following rule:

The second integral is found by first making a subsitution:

Now, rewrite the integral and solve:

We used the following rule to integrate:

Finally, replace  with our original  term and add the results of the integrations together:

To integrate, we must rewrite the integral into the sum of two integrals:

The first integral is equal to

and was found using the rule identical to the integral itself.

The second integral must be solved using integration by parts, which is given by

We are going to let

which means

The derivative and integral were found using the following rules:

, 

Now, use the above formula and integrate:

The integral was performed using the same rule as above.

Putting everything together, we get our final answer:

The integration of the function is easiest performed when you sum two separate integrals:

The first integral is equal to

and was found using the following rule:

The second integration is done by first making a substitution:

Now, rewrite the second integral and solve:

The integral was found using the following rule:

Now, replace u with the original term and add the two integrals together to get

This follows the basic intergral form for the Inverse Tangent. The form is

,where is a constant.

If you don't immediately see how matches the Inverse Tangent integral, then rewrite so that only the is in the denominator.

Then rewrite the as ,and the as , so they match the and the  in the integral form.

This matches the Inverse Tangent integral exactly with "u", "a", and "du" as follows

Now from the basic integral form, we can simply plug in "u" and "a".

This gives the correct answer of

To evaluate the integral, we must integrate by parts:

So, we must assign  and :

Next, we derive and integrate:

The derivative and integral were performed using the following rules:

, 

Next, use the above formula and integrate:

The following rule was used to integrate:

In order to calculate the indefinite integral, we apply the trigonometric property which states

As such, the indefinite integral is

In order to calculate the indefinite integral, we apply the inverse power rule which states

Applying this to the problem in this question we get

As such the indefinite integral is