# Thread: More partial fractions fun

1. ## More partial fractions fun

in the case of INTEGRAL OF: (x^3 + 7) / (x^2 + 4)

If you do long division of polynomials you get:

INTEGRAL OF: X + (-4x + 7) / (x^2 + 4)

I tried doing partial fractions here, but it doesn't work. The problem i run into is that i only have one term in my denominator, it cannot be broken up. Therefore i cant do any of the 4 cases.

I thought of something new:

I broke the fraction down into:
(-4x) / (x^2 + 4) and (7) / (x^2 + 4)

The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
I determined that the integral of that was 7( -x^[-1] + .25x)
Then i set this aside for later, knowing I will eventually have to add it to my other integral.

Next i looked at (-4x) / (x^2 + 4)
I pulled a -2 out and got then it looked like:
-2 INTEGRAL OF: ( 2x / ((x^2) + 4)
I decided to do u substitution for u = x^2 + 4 and my du was just 2xdx making dx=du/2x
So when i plug in u i get:

-2 INTEGRAL of: (2x / u) * (1/2x)du
The 2x's cancel out and your left with -2 *INTEGRAL of: 1/u du
This is clearly -2ln(u). I plug back in u and get -2ln(x^2 + 4)

So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x

But this is wrong, or so says my book.

2. Originally Posted by fogel1497
in the case of INTEGRAL OF: (x^3 + 7) / (x^2 + 4)

If you do long division of polynomials you get:

INTEGRAL OF: X + (-4x + 7) / (x^2 + 4)

I tried doing partial fractions here, but it doesn't work. The problem i run into is that i only have one term in my denominator, it cannot be broken up. Therefore i cant do any of the 4 cases.

I thought of something new:

I broke the fraction down into:
(-4x) / (x^2 + 4) and (7) / (x^2 + 4) Mr F says: Good.

The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
I determined that the integral of that was 7( -x^[-1] + .25x) Mr F says: Not so good. See below.

Then i set this aside for later, knowing I will eventually have to add it to my other integral.

Next i looked at (-4x) / (x^2 + 4)
I pulled a -2 out and got then it looked like:
-2 INTEGRAL OF: ( 2x / ((x^2) + 4)
I decided to do u substitution for u = x^2 + 4 and my du was just 2xdx making dx=du/2x
So when i plug in u i get:

-2 INTEGRAL of: (2x / u) * (1/2x)du
The 2x's cancel out and your left with -2 *INTEGRAL of: 1/u du
This is clearly -2ln(u). I plug back in u and get -2ln(x^2 + 4) Mr F says: Good.

So my final answer is .5x^2 - 2ln(x^2 + 4) - 7/x + 7/4x Mr F says: Not so good.

But this is wrong, or so says my book.
Standard form: $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C$

3. Originally Posted by fogel1497
The second of which i can integrate by pulling the 7 out and integrating x^(-2) + 1/4
I determined that the integral of that was 7( -x^[-1] + .25x)
Then i set this aside for later, knowing I will eventually have to add it to my other integral.
This is where you went wrong. Firstly:
$\frac{a}{b+c} \neq \frac{a}{b} + \frac{a}{c}$

$\frac{1}{x^2 + 4} \neq x^{-2} + \frac{1}{4}$

What you should do here is pull 4 common factor to get:

$\frac{7}{4} \int \frac{1}{(\frac{x}{2})^2 + 1}$

You should recognize this right away as a standard form (arctan).

4. $
(x^3+7)
x^2+4) = x-\frac{4x-7}{x^2+4}
" alt="
(x^3+7)x^2+4) = x-\frac{4x-7}{x^2+4}
" />
$
\int{xdx} - 4\int{\frac{xdx}{x^2+4}} + 7\int{\frac{dx}{x^2+4}}
$

$
\frac{x^2}{2}-2\ln{|x^2+4|}+\frac{7}{2}\arctan{\frac{x}{2}}+c
$