Integral of 2x/(x^2-x-6)?

I've gotten the solution from Wolfram Alpha but it uses the "completing the square" technique and the hyperbolic tangent which I didn't even know existed (these methods are not part of my course's program and it's a problem from a test).

Can anybody solve this using integration by parts or substitution? Thanks.

1 Answer
Guillaume L.
Jun 3, 2018

#int(2x)/(x²-x-6)dx=4/5ln|x-2|+6/5ln|x+3|+C#,#C in RR#

Explanation:

#int(2x)/(x²-x-6)dx#
#=2intx/(x²-x-6)dx#
#=2intx/((x+2)(x-3))dx#
Let #x/(x²-x-6)=A/(x+2)+B/(x-3)#
#x/(cancel(x²-x-6))=(Ax-3A+Bx+2B)/(cancel((x-2)(x+3)))#
#x=(A+B)x+3A-2B#
We can identify that :
#A+B=1#(1)
#2B-3A=0#(2)
(2)+3(1):
#A+B=1#
#5B=3<=>B=3/5#
So:
#A=2/5#
#B=3/5#
We have now :
#2intx/(x²-x-6)dx= 2int(2/(5(x+2))+3/(5(x-3)))dx#
#=2int(2/(5(x+2)))dx+2int(3/(5(x-3)))dx#
#=4/5int1/(x+2)dx+6/5int1/(x-3)dx#
#=4/5ln|x+2|+6/5ln|x-3|+C#,#C in RR#
\0/ here's our answer !

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