How do you evaluate the integral #int (x+2)/(x^2-2x-3)#?

Question

6886 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T14:31:15

The answer is #=-1/4ln(|x+1|)+5/4ln(|x-3|)+C#

Let's factorise the denominator

#x^2-2x-3=(x+1)(x-3)#

Let's perform the decomposition into partial fractions

#(x+2)/((x+1)(x-3))=A/(x+1)+B/(x-3)#

#=(A(x-3)+B(x+1))/((x+1)(x-3))#

The denominators are the same, we can equalize the numerators

#(x+2)=A(x-3)+B(x+1)#

Let #x=-1#, #=>#, #1=-4A#, #=>#, #A=-1/4#

Let #x=3#, #=>#, #5=4B#, #B=5/4#

Therefore,

#(x+2)/((x+1)(x-3))=(-1/4)/(x+1)+(5/4)/(x-3)#

#int((x+2)dx)/((x+1)(x-3))=-1/4intdx/(x+1)+5/4intdx/(x-3)#

#=-1/4ln(|x+1|)+5/4ln(|x-3|)+C#