Integrate xdx/((l + d -x)

You're supposed to get

- (d + l)ln( -d -l + x) -x.

Please integrate this step by step.

Assuming L and d are constants, treat them as numbers,
then x/(L+d-x) is an improper fraction,

So simplify by long division:

............_-1___
-x+L+d)x
............x-L-d
............---------
L+d

=-1+(L+d)/(-x+L+d)

So ∫-1+(L+d)/(-x+L+d) dx= -x + - ∫(L+d)/(x-L-d))dx

= -x- (L+d) ∫ 1/(x-L-d) dx

=-x-(L+d)ln(x-L-d) +C

Hoping this helps!

Let u = l+d - x , then du = -dx and x = l+d -u

The integrand is then (l+d -u)(-du)/u and it breaks up into the integrals of -(l+d)du/u +du, which are
-(l+d)ln u + u

Substitute back for u and get -(l+d)ln (l+d-x) -x +l+d. The l+d after the -x drops out as it is not a part on the indefinite integral ( it is just some constant whose derivative is 0), so what's left is

-(l+d)ln (l+d-x) -x

Check: take the derivative --- -(l+d)(-1)/(l+d-x) -1 = (l+d-l-d+x)/(l+d-x) = x/(l+d-x)

How come the expression in brackets is different from what you have? BTW, what you have is the answer you also get from the Wolfram Integrator :-)

Are the two results the same? They both have the same derivative :-) :-), but l+d-x is not equal to -l-d+x.

True, but the point is rather subtle: the argument of the logarithm has to be positive and that can only happen if the argument is |l+d-x| (which is what it should have been written), positive for ANY value of x.

So, if x > l+d, the argument is x -(l+d). If x < l+d, the argument is l+d -x.

As I said, rather subtle.

add and subtract (-l-d)/(l+d-x).

[x/(l+d-x)] -(d+l)/(l+d-x) + (d+l)/(l+d-x).

Combining the first two, you get (x-d-l)/(l+d-x) = -1.
Combining -1 with the third
-1 + (d+l)/(l+d-x).
Integration of sum = sum of integrals: