Home > Calculus > Integral by Partial Fractions

How do you integrate #(5x^2+7x-4)/(x^3+4x^2)# using partial fractions?

Answer 1

The answer is #=1/x+2lnx+3ln(x+4)+C#

let's simplify the denominator #x^3+4x^2=x^2(x+4)# So #(x^2+7x-4)/(x^2(x+4))=A/x^2+B/x+C/(x+4)#

#:.5x^2+7x-4=A(x+4)+Bx(x+4)+Cx^2# #-4=4A##=># #A=-1# coefficients of x, #7=A+4B##=>##B=2# Coefficients of x^2, #5=B+C##=># #C=3# #:. int((x^2+7x-4)dx)/(x^2(x+4))=int-dx/x^2+int(2dx)/x+int(3dx)/(x+4)# #=1/x+2lnx+3ln(x+4)+C#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Emma Aldridge

To integrate (\frac{5x^2+7x-4}{x^3+4x^2}) using partial fractions, you would first factor the denominator, then express the rational function as a sum of simpler fractions. In this case, the denominator (x^3 + 4x^2) factors into (x^2(x + 4)). The decomposition into partial fractions would have the following form:

[\frac{5x^2+7x-4}{x^3+4x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 4}]

Then, you would solve for the unknown coefficients (A), (B), and (C) by equating coefficients. After finding the values of (A), (B), and (C), you would integrate each term separately.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.