How to calculate $\int \frac{x}{\sqrt x -2}dx$

Question

I don't know how to solve the following integral. I need some suggestions. Thank you!

$$ \int \frac{x}{\sqrt x -2}dx$$

Try the cov $x=t^2$. – Martín-Blas Pérez Pinilla Feb 04 '14 at 22:41 — Martín-Blas Pérez Pinilla, Feb 04 '14 at 22:41
cov=change of variable – Martín-Blas Pérez Pinilla Feb 04 '14 at 22:50 — Martín-Blas Pérez Pinilla, Feb 04 '14 at 22:50

score 3 · Answer 1 · answered Feb 04 '14 at 22:45

3

Let $t=\sqrt x$ so $x=t^2$ and $dx=2tdt$ hence

$$\displaystyle \int \frac{xdx}{\sqrt x -2}dx=2\int\frac{t^3}{t-2}dt$$ Can you take it from here?

answered Feb 04 '14 at 22:45

score 2 · Answer 2 · answered Feb 04 '14 at 22:48

Setting $t=\sqrt{x}$ one gets \begin{eqnarray} \int\frac{x}{\sqrt{x}-2}\,dx&=&2\int\frac{t^3}{t-2}\,dt=2\int\frac{8+t^3-8}{t-2}\,dt\\ &=&2\int\left(\frac{16}{t-2}+t^2+2t+4\right)\,dt\\ &=&32\ln|t-2|+\frac23t^3+2t^2+8t+C\\ &=&32\ln|\sqrt{x}-2|+\frac23x\sqrt{x}+2x+8\sqrt{x}+C \end{eqnarray}

score 2 · Accepted Answer · edited Feb 05 '14 at 00:27

2

$$\int \frac{x }{\sqrt x -2}dx$$ $x=t^2 \ , \ dx=2t \ dt$ $$\int\frac{2t^3 }{t-2}dt=2\int\frac{t^3-8}{t-2}dt+16\int\frac{dt}{t-2}=$$ $$=2\int\frac{(t-2)(t^2+2t+4)}{t-2}dt+16\int\frac{dt}{t-2}=$$ $$=2\int(t^2+2t+4)dt+16\int\frac{dt}{t-2}=$$

edited Feb 05 '14 at 00:27

colormegone

10,842

answered Feb 04 '14 at 22:53

Adi Dani

16,949

How to calculate $\int \frac{x}{\sqrt x -2}dx$

3 Answers3