Indef. Integral

[yourmom717]

Find the indefinite integral:

(x^(1/2)/((x^(1/2)-3)

It is the squre root of x over the square root of x minus 3. I dont know how to do it.

 

[Daniel_Feldman]

let u=root(x)-3. Then root(x)=u+3. Work from there. If you get stuck, show what you've done.

 

[yourmom717]

more help

I can get that far, i just dont know what to do after that.

 

[Daniel_Feldman]

How do you normally integrate using a u-substitution?


You find dx in terms of du!

We have root(x)=u+3, so x=(u+3)^2, and dx=2(u+3)du. Now substitute, expand and integrate.

 

[soroban]

Hello, Ryan!

It would have saved time if you had shown how far you got . . .

\(\displaystyle \L\int \frac{\sqrt{x}}{\sqrt{x}\,-\,3}\,dx\)

Let \(\displaystyle u\,=\,\sqrt{x}\,-\,3\;\;\Rightarrow\;\;\sqrt{x}\,=\,u\,+\,3\;\;\Rightarrow\;\;x\,=\,(u\,+\,3)^2\;\;\Rightarrow\;\;dx\,=\,2(u\,+\,3)\,du\)

Substitute: \(\displaystyle \L\:\int\frac{u\,+\,3}{u}\cdot2(u\,+\,3)\,du \;= \;2\int\frac{(u\,+\,3)^2}{u}\,du \;= \;2\int\frac{u^2\,+\,6u\,+\,9}{u}\,du\)

If you got this far, you're doing fine . . .

Try this . . . \(\displaystyle \L2\int\left(\frac{u^2}{u}\,+\,\frac{6u}{u}\,+\,\frac{9}{u}\right)\,du \;= \;2\int\left(u\,+\,6\,+\,\frac{9}{u}\right)\,du\)

and we get: \(\displaystyle \L\:2\left(\frac{u^2}{2}\,+\,6u\,+\,9\cdot\ln|u|\right)\,+\,C\) . . . then back-substitute.