MDoubt No 000006 - GATE

GATE [EC, 2017, 2 marks]
Question :
The values of the integrals $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dy\right)dx$ and $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dx\right)dy$ are
(a) same and equal to $0.5$
(b) same and equal to $-0.5$
(c) $0.5$ and $-0.5$ respectively
(d) $-0.5$ and $0.5$ respectively

Solution : $\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dy\right)dx$
$\displaystyle=\int\limits_0^1\left[\int\limits_0^1\left(\dfrac{2x}{(x+y)^3}-\dfrac{1}{(x+y)^2}\right)dy\right]dx$
$\displaystyle=\int\limits_0^1\left[\dfrac{-x}{(x+y)^2}+\dfrac{1}{x+y}\right]_0^1dx$
$\displaystyle=\int\limits_0^1\dfrac{1}{(x+1)^2}dx$
$=\left[\dfrac{-1}{x+1}\right]_0^1$
$=-\dfrac{1}{2}+1$
$=\dfrac{1}{2}=0.5$

Now
$\displaystyle\int\limits_0^1\left(\int\limits_0^1\frac{x-y}{(x+y)^3}dx\right)dy$
$\displaystyle=\int\limits_0^1\left[\int\limits_0^1\left(\dfrac{1}{(x+y)^2}-\dfrac{2y}{(x+y)^3}\right)dx\right]dy$
$\displaystyle=\int\limits_0^1\left[\dfrac{-1}{x+y}+\dfrac{y}{(x+y)^2}\right]_0^1dy$
$\displaystyle=\int\limits_0^1\dfrac{-1}{(y+1)^2}dy$
$=\left[\dfrac{1}{y+1}\right]_0^1$
$=\dfrac{1}{2}-1$
$=-\dfrac{1}{2}=-0.5$
CORRECT ANSWER : C