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#