integrate partial fractions?

Question

help with solving 6x+14/(x-3)(x+5). I need to know the steps please. Here is anotherone 10x + 4/(x+1)^2

Answer 1

Integral [ (6x + 14) / (x - 3)(x + 5) ] dx

The partial fraction decomposition would be this:

(6x + 14) / (x - 3)(x + 5) = A/(x - 3) + B/(x + 5)

Therefore, our new integral is

Integral ( [A/(x - 3) + B/(x + 5)] dx )

For some constants A and B.
These are both easy integrals, and our solution is

A ln|x - 3| + B ln|x + 5| + C

To solve for A and B, we refer back to this equation:
(6x + 14) / (x - 3)(x + 5) = A/(x - 3) + B/(x + 5)

Multiply both sides by (x - 3)(x + 5), to get:

6x + 14 = A(x + 5) + B(x - 3)

Remember this equation is true _for all x_. Therefore, if we
let x = -5:
6(-5) + 14 = A(0) + B(-5 - 3)
-30 + 14 = B(-8)
-16 = B(-8); therefore, B = 2.

Since the equation is true for all x, let's let x = 3. Then

6(3) + 14 = A(3 + 5) + B(3 - 3)
18 + 14 = A(8) + 0
32 = 8A, so A = 4.

A = 4, B =2, and so our answer is

A ln|x - 3| + B ln|x + 5| + C before substitution, and

4 ln|x - 3| + 2 ln|x + 5| + C

Part 2:
Integral [ (10x + 4) / (x + 1)^2 dx ]

Since we have a repeated linear factor in the denominator, our partial fraction decomposition is

(10x + 4) / (x + 1)^2 = A/(x + 1) + B/(x + 1)^2

If we take the integral of this

Integral ( A/(x + 1) + B/(x + 1)^2 ) dx

We have

A ln|x + 1| - B (1/(x + 1) + C, or

A ln|x + 1| - B/(x + 1) + C

Now we solve for A and B.

(10x + 4) / (x + 1)^2 = A/(x + 1) + B/(x + 1)^2

Multiply both sides by (x + 1)^2, to get

10x + 4 = A(x + 1) + B

Let x = -1; then
10(-1) + 4 = A(0) + B
-10 + 4 = B, so B = -6.

Let x = 0; then
10(0) + 4 = A(0 + 1) + B
4 = A + B. But B = -6, so
4 = A - 6. Therefore A = 10

A = 10, B = -6, and we go from here:

A ln|x + 1| - B/(x + 1) + C

To here

10 ln|x + 1| - (-6)/(x + 1) + C

The two negatives cancel out.

10 ln|x + 1| + 6/(x + 1) + C

Answer 2

The second one is easy. You don't need partial fractions.
Just write it as 10x + 4(x+1)^(-2)
The antiderivative is 5x^2 -4/(x+1) + C
by the power rule.

For the first one write
14/(x-3)(x-5) = A/(x-3) + B/(x-5)
Now clear denominators:
14 = A(x-5) + B(x-3)
Since this holds for all values of x, we can let x = 5
to get
14 = 2B, B = 7.
Similarly, we can let x = 3 and we get
14 - 2A, A = -7.
So the first problem reduces to integrating
6x -7/(x-3) + 7(x-5),
which yields
3x² -7 ln(x-3) + 7 ln(x-5) + C.
BTW, have I misinterpreted this?
Do you want (6x+14)/(x-3)(x-5) or 6x + 14/(x-3)(x-5)
for the first one? (10x+4)/ (x+1)² or 10x + 4/(x+1)²
for the second one?

Answer 3

Here is a website that is only for partial fractions that give you step by step examples to help you.

http://www.karlscalculus.org/calc11_5.html

Hope that helps. =)