# Solve the integral using partial fraction decomposition. Show all algebraic steps and show how the constants are determined in the partial fraction decomposition. `int(3x^2+x+4)/(x^3+x)dx` (I...

Solve the integral using **partial fraction decomposition.** Show all algebraic steps and show **how the constants are determined** in the partial fraction decomposition. `int(3x^2+x+4)/(x^3+x)dx` (I am new to this partial decomposition, so i would love an explanation of how exactly to solve this.)Thanks!

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I highly recommend checking out the link I posted with this answer. That will give you a great introduction to partial fractions.

I will first explain how to set up this integral using partial fractions.

The denominator can be factored by taking an x out of each term.

`x^3 + x = x(x^2 + 1)`

These two terms, x and x^2 + 1 will become the denominators for each of our partial fractions.

This gives us:

`A/x + (Bx + C)/(x^2+1)`

The reason that we have Bx+C in the second term is that, when you have a polynomial that cannot be factored, we have to make the numerator a polynomial of degree -1 that of the denominator. (check out the table in the link I posted, or just google partial fractions table)

Now, we know that to combine these, we will multiply the first term by

`(x^2+1)/(x^2+1)` and the second term by `x/x`

This gives us `A(x^2+1)`

and `(Bx+C)x`

Now we need to find values for A, B and C so when combined it equals our numerator.

`A(x^2+1) + (Bx+C)x = 3x^2 + x + 4`

The left side simplifies to

`Ax^2 + A + Bx^2 + Cx = 3x^2 + x + 4`

Now we match things on the left side to the right side.

The is only one term on the left without an x, attached. So that means that A has to equal 4. And there is only one term on the left with a single x attached, so that means that C = 1.

This gives us

`4x^2 + Bx^2 + x + 4 = 3x^2 + x + 4`

Now we have to find a value for B. And the only value for B that works is -1.

Now we know that A = 4, B = -1, and C = 1

We can finally rewrite our integral as

`int 4/x + (-x+1)/(x^2 + 1) dx`

Now all we have left to do is integrate. I am going to let you try that on your own. I hope this help the partial fractions expansion.

**Sources:**