There are no substitutions that would work in this instance. Therefore, you must integrate by parts. The formula for integrating by parts is:

TO begin, you must assign values of u and dv based from the two different functions in the original integral. Specifically, .

From here, you can find values for du and v:

 was found by using the following formula:

V was found by using a u-substitution where  and the fact that: 

Following the formula, the integral can be rewritten as:

Once again, there is another integral that cannot be evaluated by using a substitution. Therefore, integration by parts must be used again.

The integral can be rewritten again as:

The integral that is left is easily solvable by using a u-substitution where  and the fact that .  This leaves you with:

Now, the final answer can be written as:

Despite the intimidating look of the problem, simplifications can be made. Since, the integrand can be rewritten as:

 Because of the properties of logarithms where , the integrand can be rewritten as:

There are no substitutions that can be done here. Therefore, you must integrate by parts. The formula for integrating by parts is:

To begin, you must assign values of u and dv based from the two different functions in the original integral. Specifically, .

From here, you can find values for du and v:

 was found by using the following formula:

V was found by using the following formula: 

Following the formula, the integral can be rewritten as:

The integral can be taken by using the following formula:

The final answer is:

No substitution would work here, but rewriting the integrand via long division is easiest way to solve this problem. 

After long division, the quotient should be  with a remainder of 3. Whenever there is a remainder, it must be placed over the divisor as a numerator. This makes the original integral expressed as:

Each term of the integrand can be integrated by using the following formula:

For the last term, however, a u-substitution is needed where . The integrand can be rewritten as:

The final answer is:

First, chop up the fraction into three separate terms:

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

Remember to add a C because it is an indefinite integral:

Remember that when integrating, raise the exponent by 1 and also put that result on the denominator:

Simplify to get:

Add C because it is an indefinite integral:

Using integration by parts, let , , , and .

Use the formula to get .

Then, you can use substitution for the integral in the new equation. If , then . 

Using substitution, the second half of the expression becomes, .

If you use the Power Rule, 

If you substitute this back in, you get , and you can simplify this to .

Splitting up the integral with the Sum Rule turns the problem into:.

We can then use the Power Rule to solve the equation:

, which can be simplified to .

First, FOIL the products so we have: .

Then, we can just use the Power Rule:

.

To solve the equation, one should know that the integral of  is , and the integral of . Using these rules and pulling out the constant, we get   .

To solve the indefinite integral

we use u-substitution, setting . We derive that equation to get  and so the integral becomes

The integral then becomes

and substituting back in for  yields