Integrate x/(x-1)(x^2+4) dx ?

1 Answer
Guillaume L.
Jun 25, 2018

#intx/((x-1)(x²+4))dx=1/5ln|x-1|-1/10ln(x²+4)+2/5tan^(-1)(x/2)+C#, #C in RR#

Explanation:

#intx/((x-1)(x²+4))dx#
Let #x/((x-1)(x²+4))=A/(x-1)+(Bx+C)/(x²+4)#
So:
#x=(A+B)x²+(C-B)x+4A-C#
By identification :
#A+B=0#[1]
# C-B=1#[2]
# 4A-C=0 #[3]

[3]#<-#[3]+[2]+[1]

#A+B=0#[1]
#C-B=1#[2]
#5A=1#[3]
#<=>#
#B=-1/5#
#C=4/5#
#A=1/5#

So:
#intx/((x-1)(x²+4))dx=int((1/5)/(x-1)+(-x+4)/(5(x²+4)))dx#
#=1/5int1/(x-1)dx-1/5int(x-4)/(x²+4)dx#
#=1/5ln|x-1|-1/10int(2x)/(x²+4)dx+4/5int1/(x²+4)dx#
#=1/5ln|x-1|-1/10ln(x²+4)+2/5tan^(-1)(x/2)+C#, #C in RR#
\0/ here's our answer !

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