1-calculate-dx-2-x-1-2-2-find-the-value-of-2-dx-2-x-1-2-

Question Number 94907 by mathmax by abdo last updated on 21/May/20

1) calculate \int \frac{dx}{{(2 + \sqrt{x - 1})}^{2}}

2) find the value of \int_{2}^{+ \infty} \frac{dx}{{(2 + \sqrt{x - 1})}^{2}}

Answered by MJS last updated on 22/May/20

\int \frac{d x}{{(2 + \sqrt{x - 1})}^{2}} =

[t = 2 + \sqrt{x - 1} \to d x = 2 \sqrt{x - 1} d t]

= 2 \int \frac{t - 2}{t^{2}} d t = \frac{4}{t} + 2 \ln t =

= \frac{4}{2 + \sqrt{x - 1}} + 2 \ln (2 + \sqrt{x - 1}) + C

\underset{2}{\int^{\infty}} \frac{d x}{{(2 + \sqrt{x - 1})}^{2}} diverges

Commented by mathmax by abdo last updated on 22/May/20

thank you sir mjs

Answered by mathmax by abdo last updated on 22/May/20

1) I = \int \frac{dx}{{(2 + \sqrt{x - 1})}^{2}} changement \sqrt{x - 1} = t give x - 1 = t^{2} \Rightarrow

I = \int \frac{2 tdt}{{(2 + t)}^{2}} = \int \frac{2 tdt}{t^{2} + 4 t + 4} = \int \frac{2 t + 4 - 4}{t^{2} + 4 t + 4} dt = \int \frac{2 t + 4}{t^{2} + 4 t + 4} dt

- 4 \int \frac{dt}{t^{2} + 4 t + 4} = \ln (t^{2} + 4 t + 4) - 4 \int \frac{dt}{{(t + 2)}^{2}}

= 2 \ln ∣ t + 2 ∣ + \frac{4}{t + 2} + C = 2 \ln ∣ \sqrt{x - 1} + 2 ∣ + \frac{4}{\sqrt{x - 1} + 2} + C

2) \int_{2}^{+ \infty} \frac{dx}{{(2 + \sqrt{x - 1})}^{2}} = {[2 \ln ∣ \sqrt{x - 1} + 2 ∣ + \frac{4}{\sqrt{x - 1} + 2}]}_{2}^{+ \infty} = + \infty

so this integrale is divergente .!