Question

Question #33600

Answer 1

`1/2`.

Explanation:

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

`1/2`

Explanation:

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`