Question #33600

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Answer 1 · 2017-02-28T16:14:31

$1/2$ .

Let $I=int_1^9 1/{sqrtx(1+sqrtx)^2}dx.$

Let us subst. $sqrtx=t, or, x=t^2," so that, "dx=2tdt.$

Also, $x=1 rArr t=sqrtx=sqrt1=1," &, for "x=9, t=3.$

Hence, $I=int_1^3 (2t)/{t(1+t)^2}dt=2int_1^3 (1+t)^-2dt.$

Now, recall the following Useful Result :

$int f(x)dx=F(x)+c rArr int f(ax+b)dx=1/aF(ax+b)+c', a!=0.$

Knowing that, $int t^-2dt=-1/t+C',$ we have,

$I=2[-1/(1+t)]_1^3=-2[1/4-1/2]=-2(-1/4)=1/2.$

Enjoy Maths.!

Answer 2 · 2017-02-28T16:20:23

$1/2$

Making $y = sqrt x$ and consequently $dy = 1/2(dx)/y$ we have

$int 1/(sqrtx*(1+sqrtx)^2)dx = int (2y)/(y(1+y)^2) dy = 2int (dy)/(1+y)^2=-2/(1+y)+C$

but

$int_1^9 1/(sqrtx*(1+sqrtx)^2)dx = 2int_1^(sqrt(9)) (dy)/(1+y)^2 = 1/2$