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I am having trouble figuring out an algebraic trick to make this work

Evaluate the integral

$$\int_1^9\frac{x-1}{\sqrt{x}}dx$$

I know I can turn the integrand into $(x-1) (1/\sqrt{x})$ but I still don't know how to do products of intregrals.

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break the integrand into $\sqrt{x}-\frac{1}{\sqrt{x}}$ or substitute $x=u^2$. –  robjohn Nov 3 '11 at 21:44

2 Answers 2

Hint: break the integral into $\sqrt{x}-\frac{1}{\sqrt{x}}$ or try substituting $x=u^2$.

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I got $x^.5 - x^-/5$ and the answer is wrong. –  user138246 Nov 3 '11 at 22:40
    
You still have to integrate $\sqrt{x}-\frac{1}{\sqrt{x}}$. –  robjohn Nov 3 '11 at 22:48
    
Ah you mean find what function has that as a derivative? I forgot to. –  user138246 Nov 3 '11 at 23:58

You cannot make the integrand $x-1(1/\sqrt{x})$; you can make it $(x-1)(1/\sqrt{x})$. Those parentheses are important. However, you don’t want to do any such thing. You want to divide it out. Rewrite the integrand as $$\frac{x-1}{\sqrt{x}} = \frac{x}{\sqrt{x}}-\frac1{\sqrt x}$$ and simplify each term to a power of $x$. Then integrate term by term.

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