can you please integrate sqrt(sqrt(x)-1) dx.
Thanks :)

You can consider using the substitution u=sqrt(x), where u^2 = x,
differentiating both sides gives 2u *(du/dx) = 1 =====> dx =2u du

∫ sqrt(sqrt(x)-1) dx  therefore becomes

∫ [sqrt(u-1)]*(2u) du -------------(1)

To solve (1), we can consider integration by parts such that we integrate [sqrt(u-1)] wrt u thus giving 2/3*(u-1)^(3/2) and differentiate 2u wrt u thus giving the value of 2.

So (1) becomes

[2/3*(u-1)^(3/2)]*(2u)-  ∫ 2* [2/3*(u-1)^(3/2)] du

= 4u/3*[(u-1)^(3/2)]-  ∫ 4/3*[(u-1)^(3/2)] du

= 4u/3*[(u-1)^(3/2)]- 4/3* 2/5* [(u-1)^(5/2)] +C

= 4u/3*[(u-1)^(3/2)]- 8/15* [(u-1)^(5/2)] +C

= 4u/3*[(sqrt(x)-1)^(3/2)]- 8/15* [(sqrt(x)-1)^(5/2)] +C   (shown)

Hope this helps. Peace.  


All Answers

Answers by Expert:

Ask Experts


Frederick Koh


I can answer questions concerning calculus, complex numbers, vectors, statistics , algebra and trigonometry for the O level, A level and 1st/2nd year college math/engineering student.


More than 7 years of experience helping out in various homework forums. Latest Presence is over at . You can also visit my main maths website where I have designed "question locker" vaults to store tons of fully worked math problems. A second one is currently being built. Peace.

IEEE(Institute of Electrical and Electronics Engineers )

Former straight As A level student from HCJC (aka HCI); scored distinctions in both C and Further Mathematics B Eng (Hons) From The National University Of Singapore (NUS) B Sc (Hons) From University of London External (Grad Route)

©2016 All rights reserved.