Question

Evaluate using partial fractions $\int \frac{8x-36}{(x-5{)}^2}dx$

• A. $8\ln |x-5|-\frac{4}{x-5}+c$
• B. $9\ln |x-5|-\frac{4}{x-5}+c$
• C. $8\ln |x+5|-\frac{4}{x-5}+c$
• D. $9\ln |x+5|-\frac{4}{x-5}+c$

Answer 1

The correct option is A $8\ln |x-5|-\frac{4}{x-5}+c$

Let $\frac{8x-36}{(x-5{)}^2}=\frac{A}{(x-5)}+\frac{B}{(x-5{)}^2}$ ....(1)

Multiply $(x-5{)}^2$ on both the sides, we get

$8x-36=A(x-5)+B$

$8x-36=Ax+(-5A+B)$

Comparing like terms, we get

$A=8,-5A+B=-36$

$\therefore B=4$

Substituting the values of A and B in equation (1), we get

$\frac{8x-36}{(x-5{)}^2}=\frac{8}{(x-5)}+\frac{4}{(x-5{)}^2}$

Therefore, $\int \frac{8x-36}{(x-5{)}^2}=\int \frac{8}{(x-5)}+\int \frac{4}{(x-5{)}^2}$

$=8\ln |x+5|-\frac{4}{x-5}+c$ .... Since $\int \frac{1}{(x+a{)}^2}=\frac{-1}{x+a}$