How do you integrate $\int \frac{(x - 3) (x - 1)}{- x (x - 5)} d x$ $int ((x-3)(x-1))/(-x(x-5)) dx$ using partial fractions?

Question

1385 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-18T11:53:22

$I = - x + \frac{3}{5} ln | x | - \frac{8}{5} ln | x - 5 | + C$ $I=-x+3/5ln|x|-8/5ln|x-5|+C$

Here,

$I = \int \frac{(x - 3) (x - 1)}{- x (x - 5)} d x$ $I=int((x-3)(x-1))/(-x(x-5))dx$

$= - \int \frac{x^{2} - 4 x + 3}{x^{2} - 5 x}$ $=-int(x^2-4x+3)/(x^2-5x$

$= - \int \frac{x^{2} - 5 x + x + 3}{x^{2} - 5 x}$ $=-int(x^2-5x+x+3)/(x^2-5x)$

$= - \int \frac{x^{2} - 5 x}{x^{2} - 5 x} d x - \int \frac{x + 3}{x^{2} - 5 x} d x$ $=-int(x^2-5x)/(x^2-5x)dx-int(x+3)/(x^2-5x)dx$

$= - \int 1 d x - \int \frac{x + 3}{x (x - 5)} d x$ $=-int1dx-int(x+3)/(x(x-5))dx$

$I=-x-I_1...to(D)$

We try to obtain Partial Fractions for $I_1$

$(x+3)/(x(x-5))=A/x+B/(x-5)$

$x+3=A(x-5)+Bx$

$x=0=>3=A(-5)=>A=-3/5$

$x=5=>5+3=5B=>B=8/5$

So

$I_1=int(-3/5)/xdx+int(8/5)/(x-5)dx$

$I_1=-3/5int1/xdx+8/5int1/(x-5)dx$

$I_1=-3/5ln|x|+8/5ln|x-5|+c$

Thus, from $(D)$ ,we get

$I=-x-{-3/5ln|x|+8/5ln|x-5|}+C$

$I=-x+3/5ln|x|-8/5ln|x-5|+C$

How do you integrate int ((x-3)(x-1))/(-x(x-5)) dx∫(x−3)(x−1)−x(x−5)dxint ((x-3)(x-1))/(-x(x-5)) dx using partial fractions?