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Integration by partial fractions

6. Solve for :
10. Evaluate:

11. Evaluate:

19. Express the improper rational function as the sum of a polynomial and a proper rational function:

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Integration by partial fractions

6. Solve for :
To solve this problem, we multiply both sides by dx and then integrate to get

or
.
Now we use partial fraction decomposition to separate the rational function into two more simple rational functions. Since both terms in the denominator are linear terms, we will set up the decomposition in this way:
.
Multiplying both sides of this equation by the factors and , we get

.
Condensing this equation to the first and last parts gives us

and we now set the coefficients of the x's from both sides equal to each other and the constant terms equal to each other to get the equations and . Solving the first of these two equations for A gives and substituting this into the second equation gives which has a solution of . If you multiply the top and bottom of this expression by and simplify everything, you get . Substituting this value into the equation gives . We can now substitute the values of A and B back into the integral to get
.
The constants on the tops of both of these fractions can be taken outside of the integrals and simple u substitutions will give you the solution
.
10. Evaluate:
This problem can be solved with a trigonometric substitution letting with . It may be easier to rewrite the function first to get
. Using the identity , this integral becomes
.
This integral is complicated and is generally just given from a table, but it can be solved by multiplying this by sec + tan / sec + tan in this way:

then do a u substitution with u being the denominator to get and which is what is on top of the fraction, so
. From this point, we draw a right triangle and label it from our original substitution .

From this triangle we see the relationship that . Using this and , the solution becomes
.
11. Evaluate:
This problem can also be solved with the same trigonometric substitution with and . This gives

after using the same trigonometric identity as before and cancelling one of the secant terms. If you write secant and tangent in terms of sine and cosine (sec = 1 / cos, tan = sin / cos), you can cancel the cosine terms in the denominator leaving 1 / sin in the integral where 1 / sin = csc. So, the integral becomes
.
At this point, solving the integral is similar to the integral from above. Here we multiply and divide by csc + cot in this way:
.
Then we let from the bottom and . Here we take the negative out in front of the integral and the integral becomes
.
Here we draw a right triangle again which is the same as the triangle above since our original substitution was .

From this we get and making the final solution
.

19. Express the improper rational function as the sum of a polynomial and a proper rational function:
This can be done using long division. .
Other examples of long division with polynomials can be seen at
http://www.purplemath.com/modules/polydiv2.htm
and it will show you step-by-step what is happening. Here, the resulting polynomial is and the remainder is x, so we put the remainder back over the divisor to get

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