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.......
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.
... ∫ [ x / ( 1 + d - x ) ] dx
= ∫ { [ ( 1 + d ) - ( 1 + d - x ) ] / ( 1 + d - x ) } dx
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Note this step and then Split.
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= ∫ { [ (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.
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