Since the denominator is already factored, all we need to do partial fractions is solve for the constants:
(3x^2-x)/((x^2+2)(x-3)(x-7))=(Ax+B)/(x^2+2)+C/(x-3)+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=(3(3)^2-3)/((3^2+2)(text(////))(3-7))=-6/11
We can do the same for D:
D=(3(7)^2-7)/((7^2+2)(7-3)(text(////)))=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:
3x^2-x=(Ax+B)(x-3)(x-7)-6/11(x^2+2)(x-7)+35/51(x^2+2)(x-3)
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:
click here
A=-79/561
B=-94/561
This gives that our integral is:
int\ 35/(51(x-7))-6/(11(x-3))-(79x+94)/(561(x^2+2))\ dx
The first two can be solved using rather simple u-substitutions of the denominators:
35/51ln|x-7|-6/11ln|x-3|-1/561int\ (79x)/(x^2+2)+94/(x^2+2)\ dx
We can split the remaining integral into two:
int\ (79x)/(x^2+2)+94/(x^2+2)\ dx=int\ (79x)/(x^2+2)\ dx+int\ 94/(x^2+2)\ 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 2x, so we divide by 2x to integrate with respect to u:
79int\ x/(x^2+2)\ dx=79int\ cancel(x)/(2cancel(x)u)\ du=79/2int\ 1/u\ du=79/2ln|u|+C=79/2ln|x^2+2|+C
Integral 2
We want to get this integral into the form for tan^-1:
int\ 1/(1+t^2)\ dt=tan^-1(t)+C
If we introduce a substitution with x=sqrt2u, we will be able to transform our integral into this form. To integrate with respect to u, we have to multiply by sqrt2 (since we took the derivative with respect to u instead of x):
94int\ 1/(x^2+2)\ dx=94sqrt2int\ 1/((sqrt2u)^2+2)\ du=
=94sqrt2int\ 1/(2u^2+2)\ du=94/2sqrt2int\ 1/(u^2+1)\ du=
=47sqrt2tan^-1(u)+C=47sqrt2tan^-1(x/sqrt2)+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:
35/51ln|x-7|-6/11ln|x-3|-1/561(79/2ln(x^2+2)+47sqrt2tan^-1((sqrt2x)/2))+C
