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

Question

1233 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-18T11:53:22

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

Here,

#I=int((x-3)(x-1))/(-x(x-5))dx#

#=-int(x^2-4x+3)/(x^2-5x#

#=-int(x^2-5x+x+3)/(x^2-5x)#

#=-int(x^2-5x)/(x^2-5x)dx-int(x+3)/(x^2-5x)dx#

#=-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#