Easy Indefinite Integral

• Apr 14th 2013, 02:18 PM
minneola24
Easy Indefinite Integral
Hello,
how do I find the indefinite integral? What do I set to be equal to u?

Thanks
• Apr 14th 2013, 04:53 PM
Prove It
Re: Easy Indefinite Integral
You shouldn't use substitution here.

\displaystyle \displaystyle \begin{align*} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= \int{\frac{x^2 + 5x - 8}{x^{\frac{1}{2}}}\,dx} \\ &= \int{x^{2 - \frac{1}{2}} + 5x^{1 - \frac{1}{2}} - 8x^{-\frac{1}{2}} \,dx} \\ &= \int{x^{\frac{3}{2}} + 5x^{\frac{1}{2}} - 8x^{-\frac{1}{2}}\,dx} \end{align*}

You can now use the power rule on each term.

If you REALLY want to use substitution, rewrite the integral as \displaystyle \displaystyle \begin{align*} \int{\frac{x^2 + 5x - 8}{\sqrt{x}}\,dx} &= 2\int{\frac{\left( \sqrt{x} \right) ^4 + 5 \left( \sqrt{x} \right) ^2 - 8}{2\sqrt{x}}\,dx} \end{align*} and then let $\displaystyle \displaystyle u = \sqrt{x} \implies du = \frac{1}{2\sqrt{x}}\,dx$ and the integral becomes $\displaystyle \displaystyle 2\int{u^4 + 5u^2 - 8 \, du}$.

I'm sure whichever method you choose you can now finish the problem off.
• Apr 14th 2013, 04:55 PM
iragequit
Re: Easy Indefinite Integral
complete the square then use trig substitution

oh ^ works too