How to integrate $\frac{1}{x\sqrt{x}}$

Question

How to integrate $$\frac{1}{x\sqrt{x}}$$ I don't see how could I use u substitution or integration by parts. I tried both, but it just got worse(more complex). I haven't integrate in years and I just can't warp my head around this.

Edit: Thank you everybody who helped me. It was really simple and obvious. Now it is easy.

Please read this tutorial on how to format mathematics on this site. — N. F. Taussig, Mar 18 '16 at 21:36
Which one should I accept as an answer? I guess the first in chronological order. — Szőke Szabolcs, Mar 18 '16 at 21:38
Suppose that you wanted to use a substitution (not the easiest way, but for academic interest...). A substitution like $u^2=x, 2u\text du = \text dx$ is one way, but maybe more interesting would be $v=\frac 1{\sqrt x},\text dv=-\frac{\text dx}{2x\sqrt x}$: $$\int{1\over x\sqrt x}\text dx=-2\int \text dv=-2v+C=-\frac 2{\sqrt x}+C$$ — abiessu, Mar 18 '16 at 21:41

score 4 · Answer 1 · answered Mar 18 '16 at 21:30

4

Hint: $$ \frac{1}{x\sqrt{x}}=\frac{1}{x^{3/2}}=x^{-3/2} $$

now can you find a primitive?

answered Mar 18 '16 at 21:30

Emilio Novati

62,675

score 3 · Accepted Answer · answered Mar 18 '16 at 21:30

3

$$\int\frac{dx}{x\sqrt x}=\int\frac{dx}{x^{3/2}}=\int x^{-3/2}dx=\color{red}{-\frac{2}{\sqrt x}+\mathcal C}$$

answered Mar 18 '16 at 21:30

3SAT

7,512

Tsemo Aristide · Answer 3 · 2016-03-18T21:32:29.123

2

Hint: $1/(x\sqrt x) = x^{-3/2}$ so its primitive is $-2x^{-1/2}$.

edited Mar 18 '16 at 21:32

answered Mar 18 '16 at 21:29

Tsemo Aristide

87,475

3

That is more than a hint. – N. F. Taussig Mar 18 '16 at 21:32

Jan Eerland · Answer 4 · 2016-03-18T21:42:09.293

1

Notice, when $n>1$:

$$\int\frac{1}{x^n}\space\text{d}x=\int x^{-n}\space\text{d}x=\frac{x^{1-n}}{1-n}+\text{C}$$

edited Mar 18 '16 at 21:42

answered Mar 18 '16 at 21:36

Jan Eerland

28,671

Yes. I've used this. Thanks. I just overlooked the fact that I can move x inside the √ as x² – Szőke Szabolcs Mar 18 '16 at 21:39
@SzőkeSzabolcs You're welcome! – Jan Eerland Mar 18 '16 at 21:40
Should be OK over any interval with $x > 0$, but the final expression won't be valid if $n=1$. – Mar 18 '16 at 21:41
@Bungo Yes that's true, thats why I choose to put in $n>1$ because the antiderivative of $\frac{1}{x}$ is $\ln|x|+\text{C}$ – Jan Eerland Mar 18 '16 at 21:42

How to integrate $\frac{1}{x\sqrt{x}}$

4 Answers4