# How do you integrate f(x)=(3x^2-x)/((x^2+2)(x-3)(x-7)) using partial fractions?

Feb 7, 2018

$\frac{35}{51} \ln | x - 7 | - \frac{6}{11} \ln | x - 3 | - \frac{1}{561} \left(\frac{79}{2} \ln \left({x}^{2} + 2\right) + 47 \sqrt{2} {\tan}^{-} 1 \left(\frac{\sqrt{2} x}{2}\right)\right) + C$

#### Explanation:

Since the denominator is already factored, all we need to do partial fractions is solve for the constants:
$\frac{3 {x}^{2} - x}{\left({x}^{2} + 2\right) \left(x - 3\right) \left(x - 7\right)} = \frac{A x + B}{{x}^{2} + 2} + \frac{C}{x - 3} + \frac{D}{x - 7}$

Note that we need both an $x$ and a constant term on the left most fraction because the numerator is always of 1 degree lower than the denominator.

We could multiply through by the left hand side denominator, but that would be a huge amount of work, so we can instead be smart and use the cover-up method.

I won't go over the process in detail, but essentially what we do is find out what makes the denominator equal zero (in the case of $C$ it is $x = 3$), and plugging it into the left hand side and evaluating while covering up the factor corresponding to the constant this gives:
$C = \frac{3 {\left(3\right)}^{2} - 3}{\left({3}^{2} + 2\right) \left(\textrm{/ /}\right) \left(3 - 7\right)} = - \frac{6}{11}$

We can do the same for $D$:
$D = \frac{3 {\left(7\right)}^{2} - 7}{\left({7}^{2} + 2\right) \left(7 - 3\right) \left(\textrm{/ /}\right)} = \frac{35}{51}$

The cover-up method only works for linear factors, so we're forced to solve for the $A$ and $B$ using the traditional method and multiplying through by the left hand side denominator:
$3 {x}^{2} - x = \left(A x + B\right) \left(x - 3\right) \left(x - 7\right) - \frac{6}{11} \left({x}^{2} + 2\right) \left(x - 7\right) + \frac{35}{51} \left({x}^{2} + 2\right) \left(x - 3\right)$

If we multiply through all of the parenthesis and equate all the coefficients of the various $x$ and constant terms, we can find out the values of $A$ and $B$. It is a rather lengthy calculation, so I will just leave a link for whoever is interested:
$A = - \frac{79}{561}$
$B = - \frac{94}{561}$

This gives that our integral is:
$\int \setminus \frac{35}{51 \left(x - 7\right)} - \frac{6}{11 \left(x - 3\right)} - \frac{79 x + 94}{561 \left({x}^{2} + 2\right)} \setminus \mathrm{dx}$

The first two can be solved using rather simple u-substitutions of the denominators:
$\frac{35}{51} \ln | x - 7 | - \frac{6}{11} \ln | x - 3 | - \frac{1}{561} \int \setminus \frac{79 x}{{x}^{2} + 2} + \frac{94}{{x}^{2} + 2} \setminus \mathrm{dx}$

We can split the remaining integral into two:
$\int \setminus \frac{79 x}{{x}^{2} + 2} + \frac{94}{{x}^{2} + 2} \setminus \mathrm{dx} = \int \setminus \frac{79 x}{{x}^{2} + 2} \setminus \mathrm{dx} + \int \setminus \frac{94}{{x}^{2} + 2} \setminus \mathrm{dx}$

I will call the left one Integral 1 and the right one Integral 2.

Integral 1
We can solve this integral by a u-substitution of $u = {x}^{2} + 2$. The derivative is $2 x$, so we divide by $2 x$ to integrate with respect to $u$:
$79 \int \setminus \frac{x}{{x}^{2} + 2} \setminus \mathrm{dx} = 79 \int \setminus \frac{\cancel{x}}{2 \cancel{x} u} \setminus \mathrm{du} = \frac{79}{2} \int \setminus \frac{1}{u} \setminus \mathrm{du} = \frac{79}{2} \ln | u | + C = \frac{79}{2} \ln | {x}^{2} + 2 | + C$

Integral 2
We want to get this integral into the form for ${\tan}^{-} 1$:
$\int \setminus \frac{1}{1 + {t}^{2}} \setminus \mathrm{dt} = {\tan}^{-} 1 \left(t\right) + C$

If we introduce a substitution with $x = \sqrt{2} u$, we will be able to transform our integral into this form. To integrate with respect to $u$, we have to multiply by $\sqrt{2}$ (since we took the derivative with respect to $u$ instead of $x$):
$94 \int \setminus \frac{1}{{x}^{2} + 2} \setminus \mathrm{dx} = 94 \sqrt{2} \int \setminus \frac{1}{{\left(\sqrt{2} u\right)}^{2} + 2} \setminus \mathrm{du} =$

$= 94 \sqrt{2} \int \setminus \frac{1}{2 {u}^{2} + 2} \setminus \mathrm{du} = \frac{94}{2} \sqrt{2} \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du} =$

$= 47 \sqrt{2} {\tan}^{-} 1 \left(u\right) + C = 47 \sqrt{2} {\tan}^{-} 1 \left(\frac{x}{\sqrt{2}}\right) + C$

Completing the original integral
Now that we know what Integral 1 and Integral 2 is equal to, we can complete the original integral to get our final answer:
$\frac{35}{51} \ln | x - 7 | - \frac{6}{11} \ln | x - 3 | - \frac{1}{561} \left(\frac{79}{2} \ln \left({x}^{2} + 2\right) + 47 \sqrt{2} {\tan}^{-} 1 \left(\frac{\sqrt{2} x}{2}\right)\right) + C$