URGENT!! What's the integral of ln(ax+b)?

Question

I know that the integral of ln(x) is xln(x)-x+c.
And the integral of ln(ax) is xln(ax)-x+c.
But How about ln(ax+b)?

10 points to the anyone who can answer this question immediately!!!

Answer 1

u = ax + b

du = a * dx



ln(ax + b) * dx =>

(1/a) * ln(u) * du



The integral of which is:



(1/a) * (u * ln(u) - u) + C =>

(1/a) * ((ax + b) * ln(ax + b) - (ax + b)) + C

Answer 2

Integral Of Ax

Answer 3

Integration by substitution is not needed here, yet several answers have used it.



Work out this integral by replacing x with ax + b and dividing by a:

∫ ln(ax + b) dx = [(ax + b)ln(ax + b) - (ax + b)] / a + C

∫ ln(ax + b) dx = [(ax + b)ln(ax + b) - ax - b] / a + C

∫ ln(ax + b) dx = (ax + b)ln(ax + b) / a - x - b / a + C

∫ ln(ax + b) dx = (ax + b)ln(ax + b) / a - x + C

Answer 4

Use substitution-integration method to find the integral of ln|ax + b|, which is ∫ln(ax + b) dx = 1/a * ((ax + b)ln(ax + b) - (ax + b)) + c



I hope this helps!

Answer 5

What do you want to integrate with respect to. The only thing I can tell you is when you integrate the ln of anything, you have to use integration by parts.

Answer 6

one million) First make u = sqrt (x). 2) Now, isolate the x by capacity of adjusting for x (this eliminates the entire sq. root factor, by way of fact elevating the sqrt of something to the 2d skill, eliminates the sq. root) so which you have: u^2 = x 3) Now, take the spinoff of the two aspects: 2u du = dx 4) Now you have dx = 2u*du something and sqrt(x) = u, so which you would be able to plug those back into the concern: S [ (one million+U)/U ] * 2u du 5) Multiply the 2u for the duration of the equation to get: S 2U + 2 6) you may now combine it actual: S 2u+2 = [ u^2 + 2u ] 7) Now remember u = sqrt(x) spectacular? So purely plut sqrt(x) back into u: u^2 + 2u = (sqrtx)^2 + 2sqrt(x) 8) Sqrt x strengthen to the 2nd skill is purely x, in an effort to simplify this slightly greater to finally get: x + 2 sqrt(x) SO the respond of S (one million+sqrt(x))/(sqrtx) = x + 2 sqrt(x)

Answer 7

integral of In(ax+b)

= (x+b/a)In(ax+b) -x+constant

Answer 8

∫ln(ax + b) dx



u = ax + b



du/a = dx



1/a*∫ln(u) du



1/a*[u*ln(u) - u] + C



1/a*[(ax + b)*ln(ax + b) - (ax + b)] + C

Answer 9

www.wolframalpha.com can solve pure computational problems if that is what you desire.

http://www.wolframalpha.com/input/?i=integral+ln%28ax%2Bb%29

Answer 10

Go to a site www.wolframalpha.com , just type your problem in

Answer 11

((ax+b)log(ax+b)-ax) /a +constant