How do you integrate #(x-5)/(x-2)^2# using partial fractions?

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Answer 1 · 2016-07-29T15:59:53

#int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C#

#int (x-5)/(x-2)^2 .d x=?#

#"let us write (x-5) as (x-2-3)"#

#int (x-2-3)/(x-2)^2 * d x#

#"split (x-2-3 )#

#int cancel((x-2))/cancel((x-2)^2)* d x-int 3/(x-2)^2 *d x#

#int (d x)/(x-2)-3 int (d x)/(x-2)^2#

#"1- solve " int (d x)/(x-2)#

#"substitute u=x-2" ; "d u=d x#

#int (d x)/(x-2)=(d u)/u=l n u#

#"undo substitution "#

#int (d x)/(x-2)=l n (x-2)#

#"2- now solve "3 int (d x)/(x-2)^2#

#u=x-2" ; " d x=d u" ; "3 int (d u)/u^2=3int u^-2 d u#

#3int (d x)/(x-2)=1/(-2+1 )u^(-2+1)=-3*u^(-1)=-3/u#

#"undo substitution"#

#3int(d x)/(x-2)^2=-3/(x-2)#

#"The problem is solved:"#

#int (x-5)/(x-2)^2 .d x=l n(|x-2|)-3/(x-2)+C#