I need a little help with an integration problem?

Question

The indefinite integral (wrt x):

1 / (bx+ax^2)

Or, if you like:

1 / [x(b+ax)]

I suspect that this is IBP's, but I think I need to perform an algebra trick first to make it work...I don't know...any ideas from those in the know?

Answer 1

You are right here
Int 1 / (bx+ax^2)

= Int 1 / [x(b+ax)]
Then use partial fractions
Let 1 / [x(b+ax)] = P/x +Q/(ax+b)

= (P(ax+b) +Qx)/(ax+b)
Pa + Q = 0 and Pb =1
P=1/b
Q = -Pa = -a/b

Int 1 / [x(b+ax)]
=int [ 1/(bx) -a/(b(ax+b))
= ln|bx|/b -a/b x ln |ax+b|/a + k
=(ln|bx| - ln |ax+b|)/b + k

Also ln bx = ln b + ln x
So we can write

(ln|bx| - ln |ax+b|)/b + k
=(ln|x| - ln |ax+b|)/b + k

Answer 2

1/x(ax+b)=[1/b1/x-a/(ax+b)]
its integral=1/b[lnx-ln(ax+b)
=1/b lnx/(ax+b).