Question

More on Repeated Linear Factors The Heaviside Method is not very effective at finding the unknown numerators for par- tial fraction decompositions with repeated linear factors, but here is another way to find them. (a) If \(\frac{x^{2}+3 x+5}{(x-1)^{3}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}},\) show that \(A(x-1)^{2}+B(x-1)+C=x^{2}+3 x+5\) (b) Expand and equate coefficients of like terms to show that \(A=1,-2 A+B=3,\) and \(A-B+C=5 .\) Then find \(A, B\) , (c) Use partial fractions to evaluate \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\)

Short answer

The values for \(A\), \(B\), and \(C\) are \(1\), \(5\), and \(9\) respectively. The integral of \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\), after substituting these values, evaluates to \(\ln|x-1| - \frac{5}{x - 1} - \frac{9}{2(x - 1)^{2}} + constant\).

Step-by-step solution

Expand the Left Hand Side of the Equation

Expand \(A(x-1)^{2}+B(x-1)+C\). First, expand \(A(x-1)^{2}\) to obtain \(Ax^{2} - 2Ax + A\). Similarly, expand \(B(x-1)\) to get \(Bx - B\). So, once simplified, the left side of the equation becomes \(Ax^{2} - 2Ax + A + Bx - B + C\).

Equate Coefficients of the Same Terms

From the left hand side of the equation we obtained in the previous step, now equate coefficients of like terms. This gives you three equations: \(A=1\) for the coefficients of \(x^{2}\), \(-2A + B = 3\) for the coefficients of \(x\) and \(A - B + C = 5\) for the constants.

Solve the System of Equations

We already have \(A = 1\) from equating the coefficients of \(x^{2}\). Substituting \(A = 1\) into \(-2A + B = 3\) gives \(B=3+2=5\). Substituting \(A = 1\) and \(B = 5\) into \(A - B + C = 5\) gives \(C=5-1+5=9\). So, \(A = 1\), \(B = 5\) and \(C = 9\).

Perform Partial Fraction Integration

Substitute the obtained values of \(A\), \(B\), and \(C\) into the integral \(\int \frac{x^{2}+3 x+5}{(x-1)^{3}} d x\) which simplifies to \(\int \left(\frac{1}{x - 1} + \frac{5}{(x - 1)^{2}} + \frac{9}{(x - 1)^{3}}\right)dx\). Now integrate term by term: \(\ln|x-1| - \frac{5}{x - 1} - \frac{9}{2(x - 1)^{2}} + constant\).