Question

$\int \frac{3x+1}{(x-1{)}^2(x+3)}dx=$

• A. $\log |\frac{x-1}{x+3}|-\frac{1}{x-1}+c$
• B. $\frac{1}{2}\log |\frac{x-1}{x+3}|-\frac{1}{x-1}+c$
• C. $\frac{1}{2}\log |\frac{x-1}{x+3}|+\frac{1}{x+1}+c$
• D. $\frac{1}{2}\log |\frac{x+1}{x-3}|+\frac{1}{x+1}+c$

Answer 1

The correct option is B $\frac{1}{2}\log |\frac{x-1}{x+3}|-\frac{1}{x-1}+c$

$\int \frac{3x+1}{(x-1{)}^2(x+3)}dx$

$=\int \frac{3x+9-8}{(x-1{)}^2(x+3)}dx$

$=\int \frac{3(x+3)}{(x-1{)}^2(x+3)}dx-\int \frac{8}{(x-1{)}^2(x+3)}$

$=3\int \frac{1}{(x-1{)}^2}dx-2\int \frac{(x+3)-(x-1)}{(x-1{)}^2(x+3)}dx$

$=3\int \frac{1}{(x-1{)}^2}dx-2\int \frac{1}{(x-1{)}^2}dx+2\int \frac{1}{(x-1)(x+3)}$

$=\int \frac{1}{(x-1{)}^2}+\frac{2}{4}[\int \frac{(x+3)-(x-1)}{(x-1)(x+3)}dx]$

$=-\frac{1}{x-1}+\frac{1}{2}\log \left|\frac{x-1}{x+3}\right|+c$

$=\frac{1}{2}\log \left|\frac{x-1}{x+3}\right|-\frac{1}{x-1}+c$