How do you integrate $\frac{x + 2}{x (x - 4)}$ $(x+2) / (x(x-4))$ using partial fractions?

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Answer 1 · 2018-07-17T21:25:03

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

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

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

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

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

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

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

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

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