Tough Integral?

Question

Integrate the function: (6 * dy) / (sqrt(y) * (1+y))

I know the answer is 12 * arctan (sqrt(y)) + C, but how?

Answer 1

Your integral equates to this one:

Integral ( 6 / [sqrt(y) (1 + y)] ) dy

First, let's pull out the constant 6.

6 * Integral (1 / [sqrt(y) (1 + y)] ) dy

Now, we use u-substitution.

Let u = sqrt(y). Then
u^2 = y.
u^2 + 1 = y + 1, and now, differentiating,

2u du = dy

Our integral then becomes

6 * Integral ( 1 / [u (u^2 + 1) ] (2u) du )

Pulling the 2 out of the constant, we have

12 * Integral ( 1 / [u(u^2 + 1)] (u) du )

Notice the u next to the du will cancel with the u in the denominator. That leaves us with

12 * Integral ( 1 / (u^2 + 1) ) du

Now, we can solve this integral directly, as 1/(u^2 + 1) is one of our known derivatives; it's the derivative of arctan.

12 * arctan(u) + C

Substituting u = sqrt(y) back, we have

12 * arctan[sqrt(y)] + C

Answer 2

(6 * dy) / (sqrt(y) * (1+y))

let==> sqrt(y) = x than: y = x^2 and dy = 2x dx
put it in the function so : (6* 2x dx) / x * (1+x^2)
integral of :12x dx /x *(1+x^2)
divide on x ===> 12 dx / (1+x^2)
putting the formula integral of dx / (1+x^2) = arctg(X)
than integral of 12 dx / (1+x^2) function is : 12 arctg (X)+C
now put the value of x i mean : sqrt(y) = x
than 12arctg[sqrt(y)] + c