Verified answer

It is an irrational function, therefore you should always start with a substitution

sqrt(x)=t

x=t^2

dx=2t*dt

So input everywhere new values, you'll get

∫t*2t*dt/(t-3) = 2∫((t^2)/(t-3)) dt

(t^2)/(t-3) is an improper fraction, therefore you have to convert it in a forb a+b/c

To do so you have to divide t^2 with (t-3) (You can use polynomial long division)

You'll get

(t^2)/(t-3) = t+3+9/(t-3)

Return to the integral:

2∫((t^2)/(t-3)) dt = 2∫( t+3+9/(t-3)) dt = 2∫tdt + 6∫dt + 18∫1/(t-3) dt=

2∫tdt + 6∫dt + 18∫1/(t-3) d(t-3)=t^2+6t+18ln|t-3|+c

Now you have to return to x, so substitute t=sqrt(x)

You'll get x+6sqrt(x)+18ln|sqrt(x)-3|+c

We use algebraic substitution.

Let u = √x

This means u = x^(1/2)

and therefore

du = (1/2)x^(-1/2) dx

du = 1 / [2x^(1/2)] dx

du = 1 / [2 √x] dx

dx = 2 √x du

dx = 2 u du

Now substitute into the original integral

∫ (√x) / (√x - 3) dx = ∫ [u / (u - 3)] 2u du

∫ (√x) / (√x - 3) dx = ∫ 2u² / (u - 3) du

Now let z = u - 3 (which means z + 3 = u)

dz = du

∫ 2u² / (u - 3) du = ∫ [2(z + 3)² / z] dz

∫ 2u² / (u - 3) du = ∫ [2(z² + 6z + 9) / z] dz

∫ 2u² / (u - 3) du = ∫ [(2z² + 12z + 18) / z dz

∫ 2u² / (u - 3) du =∫ [2z + 12 + 18/z] dz

∫ 2u² / (u - 3) du = z² + 12z + 18 ln |z| + k

∫ 2u² / (u - 3) du = (u - 3)² + 12(u - 3) + 18 ln |u - 3| + k

∫ (√x) / (√x - 3) dx = (√x - 3)² + 12(√x - 3 ) + 18 ln |√x - 3| + k

∫ (√x) / (√x - 3) dx = (√x - 3)(√x - 3) + 12(√x - 3 ) + 18 ln |√x - 3| + k

∫ (√x) / (√x - 3) dx = x - 6√x + 9 + 12√x - 36 + 18 ln |√x - 3| + k

∫ (√x) / (√x - 3) dx = x + 6√x - 27 + 18 ln |√x - 3| + k

Now -27 + k can be regarded as a new constant, say c

∫ (√x) / (√x - 3) dx = x + 6√x + 18 ln |√x - 3| + c

Phew !

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∫ √x/(√x-3) dx =

=2∫ x/(√x-3) d(√x - 3) =

Let v=√x - 3, x=(v+3)²

=2∫ (v+3)²/v dv =

=2∫ v+6+9/v dv =

=2∫ v dv +12∫ 1 dv+18∫ 1/v dv =

= v² +12v+18ln|v| +C`=

= (√x - 3)² +12(√x - 3)+18ln|√x - 3| +C`=

= x - 6√x +12√x +18ln|√x - 3| +C=

= x +6√x +18ln|√x - 3| +C

http://www.wolframalpha.com/input/?i=%E2%88%AB+%E2...