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