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# Calculus/calculus

Question
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.

Calculus

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