First factor the denominator and then write the results as a sum of two fractions as follows, 

                   (1)

Here we have applied the reverse of the cross multiplication process. Now we carefully deduce the numerators  and . Cross multiply as follows: 

                                 (2)

Looking at Equation (1) and Equation (2) it's clear that we must find values of and  such that:

                                                      (3)

Note that Equation (3) is an identity, meaning it must be true for all values of  Therefore, we can freely choose a value of  to eliminate either  or . Choosing  eliminates . 

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Alternatively,  and  could have been deduced by expanding the left side of Equation (3) and collecting like terms as follows, 

                                                        (4)

 Upon inspection of Equation (4) we see that the two following conditions for  and  are required, 

Solve for  and  , this can be done quickly by adding the two equations above, 

Using this value,  gives . 

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Summarizing our partial fraction decomposition: 

Now we can use the partial fraction decomposition to compute the integral.  

We can apply the sum rule on the right side and recognize that each term is the derivative of a natural logarithm, 

We could use the properties of logarithms to write the result as a single logarithm, but we will leave it in the form of a sum here, remembering the constant of integration . 

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Now we can check to make sure the result is correct by differentiating the result: 

 First we can start by calling the integral  for convenience in a later step.

                                                                         (1)

Let's rewrite the exponential function  using the inverse relationship between  and , 

So now Equation (1) can be written as: 

                                                            (2)

 To evaluate this integral we will use integration by parts. We start by carefully defining two new variables that will break our integral down into more manageable terms. 

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Setup I.B.P

                                                            (3) 

  and  are both functions of . 

Determining how to define the new variables  and  can be a process of trial and error. Once you define  and , you can find the other parts,  and . 

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Now we can look off Equation (3) to assemble the integral, then simplify: 

                               (4)

Notice that the second term on the right-side of Equation (4) has yet another instance where we are multiplying an exponential function by a trigonometric function, a situation very similar to the original integral Equation (2).

Now let's perform integration by parts on the second term in Equation (4), ignoring the  in front of the integral for a moment.    

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Setup I.B.P Again for 

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  Fill everything in and simplify:  

Note that the original integral  has appeared in the equation. 

                         (6)

Now let's substitute Equation (6) this into Equation (4).  

                   (7)

Now solve Equation (7) for . 

Now we can write it using , 

Finally, 

To solve this equation we must integrate by parts by using the equation

.  

The first step in integrating by parts is to define u and dv.  

In this problem  and .  

We then need to find du and v.  

Taking the derivative of u we get  and taking the integral of dv we get .  

Thus we can plug in the values and get 

For integration, we do the opposite of a derivative.  Using the power rule, we increase the exponent by one and divide by the new exponent.

.

Simplifying, we get:

.

Remember, because this is an indefinite integral, we have to add the constant at the end ().

For this problem we have to use substitution to solve the integral.  set u=,  then du=.   then we see that    .

Therefore: