Integration with partial fractions

Question

$$\int\frac{2}{x(3x-8)}dx=P\cdot \ln\left|x\right|+Q\cdot \ln\left|3x-8\right|$$

Find out what P and Q are equal to.

This is what I worked out:

$$\frac{A}{x}+\frac{B}{3x-8}=\frac{2}{x(3x-8)}$$ $$-\frac{1}{4}=A,\ \ \ \frac{3}{4}=B$$ $$P=A, Q=B$$

why is the answer $P=-\frac{1}{4}, Q=\frac{1}{4}$?

Answer 1

You have the integral $\int\dfrac{2}{x(3x-8)}=2\int\dfrac{1}{x(3x-8)}$

Now if you solve $\dfrac{1}{x(3x-8)}$ using partial fractions then you will get $$\dfrac{1}{x(3x-8)}=\frac{3}{8(3x-8)}-\frac{1}{8x}$$ Now, $$2\int\dfrac{1}{x(3x-8)}=2\int\frac{3}{8(3x-8)}-\frac{1}{8x}$$ $$=2\left(\frac18\ln|3x-8|-\frac18\ln|x|\right)=\frac14\ln|3x-8|-\frac14\ln|x|$$ Therefore, $P=-\dfrac14$ and $Q=\dfrac14$

Edit: $$2\int\frac{3}{8(3x-8)}dx=\frac34\int\frac{1}{3x-8}dx$$ Apply $u=3x-8$ and we get $$=\frac34\int\frac{1}{3u}=\frac14\int\frac1udu=\frac14\ln|u=\frac14\ln|3x-8|$$

And

$$2\int\frac{1}{8x}=\frac{2}{8}\ln|x|=\frac14\ln|x|$$

Answer 2

Since $$\frac{{\rm d}(P\cdot \ln\left|x\right|+Q\cdot \ln\left|3x-8\right|)}{{\rm d}x}=\frac{P}{x}+\frac{3Q}{3x-8}=\frac{3(P+Q)x-8P}{x(3x-8)},$$

we obtain $$P+Q=0,~~-8P=2.$$ As a result, $$P=-\frac{1}{4}, ~~~Q=\frac{1}{4}.$$