How do you integrate # (x+2) / (x(x-4))# using partial fractions?

1 Answer

#\color{red}{\int \frac{x+2}{x(x-4)}\ dx}=\color{blue}{-1/2\ln |x|+3/2\ln|x-4|+C#

Explanation:

Let

# \frac{x+2}{x(x-4)}=A/x+B/{x-4}#

#(A+B)x-4A=x+2#

Comparing the corresponding coefficients on both the sides we get

#A+B=1\ \ &\ \ -4A=2#

Solving above equations we get

#A=-1/2, B=3/2#

Now, the partial fractions can be written as follows

# \frac{x+2}{x(x-4)}=(-1/2)/x+(3/2)/{x-4}#

#=-1/{2x}+3/{2(x-4)}#

#\therefore \int \frac{x+2}{x(x-4)}\ dx#

#=\int (-1/{2x}+3/{2(x-4)})\ dx#

#=-1/2\int dx/x +3/2\int dx/{x-4}#

#=-1/2\ln |x|+3/2\ln|x-4|+C#