integrate xdx/((l + d -x)?

Question

You're supposed to get

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

Please integrate this step by step.

Answer 1

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!

Answer 2

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.

Answer 3

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:

Integ [-1 + (d+l)/(l+d-x). ]dx = Integ[-1]dx + Integ[(d+l)/(l+d-x)]dx.



Integrating the first gives you -x.

Integrating the second:

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

Double negate it : [(d+l)/(l+d-x)] = -1*[-(d+l)/(l+d-x)].

U substitution: u = l+d-x. du = -1.



the second integral becomes INTEG[ -(d+l)/u]du. = -(d+l)*INTEG[1/u]du.

= -(d+l)*ln(u) = -(d+l)*ln(l+d-x).

Add the -x and you get your final answer.

Answer 4

... ∫ [ x / ( 1 + d - x ) ] dx



= ∫ { [ ( 1 + d ) - ( 1 + d - x ) ] / ( 1 + d - x ) } dx

.........................................................................



Note this step and then Split.

..........................................................................



= ∫ { [ (1+d ) / (1+d-x) ] - 1 } dx



= (1+d) ∫ [ 1 / (1+d-x) ] dx - ∫ 1 dx



= (1+d) ∫ ( 1 / u ) ( - du ) - x, ............ u = 1+d-x



= -(1+d) ln | u | - x + C



= -(1+d)· ln | 1+d-x | - x + C ....................... Ans.

_______________________________



Happy To Help !

_______________________________