indefinate integral

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craziebbygirl

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Hi I am a little confused when you are integrating a fraction, I don't really understand how to do it is there a general rule.

I have the problem: find the integral of (5x - 6) / (square root of X) dx

Can you separate it into 5x / X^1/2 - 6/x^1/2? even if I do that I still don't see how to solve it?
 
\(\displaystyle \int\frac{5x-6}{x^{1/2}}dx = \int x^{-1/2}(5x-6)dx = \int (5x^{1/2}-6x^{-1/2})dx\)

\(\displaystyle Can \ you \ take \ it \ from \ here?\)
 
craziebbygirl said:
Hi I am a little confused when you are integrating a fraction, I don't really understand how to do it is there a general rule.

I have the problem: find the integral of (5x - 6) / (square root of X) dx

Can you separate it into 5x / X^1/2 - 6/x^1/2?

Yes you can ... then it becomes

5x[sup:29h4rcmo]1/2[/sup:29h4rcmo] - 6x[sup:29h4rcmo]-1/2[/sup:29h4rcmo]

Now these are of the for x[sup:29h4rcmo]n[/sup:29h4rcmo] which when integrated becomes x[sup:29h4rcmo]n+1[/sup:29h4rcmo]/(n+1) + C

just like Glen showed you above....

even if I do that I still don't see how to solve it?
Click to expand...
 
craziebbygirl said:
Yes I understand not the answer would be 10/3x^3/2 - 12x^1/2 + c <<<

You can differentiate to check your answer

d/dx [10/3 * x[sup:32tbh6ow]3/2[/sup:32tbh6ow] - 12 * x[sup:32tbh6ow]1/2[/sup:32tbh6ow] + c] = 10/3 * 3/2 * x[sup:32tbh6ow]1/2[/sup:32tbh6ow] - 12 * 1/2 * x[sup:32tbh6ow]-1/2[/sup:32tbh6ow] = 5 * x[sup:32tbh6ow]1/2[/sup:32tbh6ow] - 6 * x[sup:32tbh6ow]-1/2[/sup:32tbh6ow]

You must be correct....


Thanks!
Click to expand...
 
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