Question

$\int \left(\frac{1}{x-3}-\frac{1}{x^2-3x}\right)dx=$ ____________ $+c;x>3$.

• A. $\frac{2}{3}\log \left(x\right(x-3\left)\right)$
• B. $\frac{1}{3}\log \left(\sqrt{x}\right(x+3\left)\right)$
• C. $\frac{2}{3}\log \left(\sqrt{x}\right(x-3\left)\right)$
• D. $\frac{1}{3}\log \left(x\right(x-3\left)\right)$

Answer 1

The correct option is C $\frac{2}{3}\log \left(\sqrt{x}\right(x-3\left)\right)$

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

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

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

$=\log (x-3)-\frac{1}{3}\left(\log \right(x-3)-\log (x\left)\right)$

$=\frac{2}{3}\log (x-3)+\frac{1}{3}\log \left(x\right)$

$=\frac{2}{3}\log (x-3)+\frac{2}{3}\log \left(\sqrt{x}\right)$

$=\frac{2}{3}\log \left(\sqrt{x}\right(x-3\left)\right)$

Hence, the answer is option (C).