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

Question

Luke Alvarez · Accepted Answer

#35/51ln|x-7|-6/11ln|x-3|-1/561(79/2ln(x^2+2)+47sqrt2tan^-1((sqrt2x)/2))+C#

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#

Christopher Agresta · Answer

undefined To integrate $ f(x) = \frac{{3x^2 - x}}{{(x^2 + 2)(x - 3)(x - 7)}} $ using partial fractions, you first factor the denominator completely. Then, express $ f(x) $ as the sum of simpler fractions with undetermined constants. After that, equate the original expression to the sum of the partial fractions and solve for the constants. Finally, integrate each term separately.