[SOLVED] Integral and Error Bounds

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March 2nd 2009, 11:47 PM
sleepiiee

[SOLVED] Integral and Error Bounds

I can't quite found out how to integrate this
(dx)/(x + x^(1/3))
I've tried doing it by partial fraction decomposition and u-sub x^(2/3)
Can anyone guide me to the way to solve this?

Also, how do you find K when finding out error bounds for Midpoint/Trapezoid
I know your suppose to find f''(x) but I don't get what you do after that
March 3rd 2009, 12:04 AM
sleepiiee

nvm found the answers
March 3rd 2009, 12:05 AM
redsoxfan325

$\int\frac{dx}{x+x^{\frac{1}{3}}}$

Let $x = u^3$ . Thus $dx=3u^2\,du$ .

Subbing in gives: $\int\frac{3u^2}{u^3+u}\,du$ .

Factor out the $3$ and cancel out a $u$ , which leaves: $3\int\frac{u}{u^2+1}\,du$ .

Let $t=u^2+1$ . Thus $dt=2u\,du$ .

Thus the integral becomes: $\frac{3}{2}\int\frac{dt}{t}\,dt$ .

This equals $\frac{3}{2}ln(t) = \frac{3}{2}ln(u^2+1)$ .

Subbing back in $x^{\frac{1}{3}}$ for $u$ gives: $\frac{3}{2}ln((x^{\frac{1}{3}})^2+1)$ .

Thus, $\int\frac{dx}{x+x^{\frac{1}{3}}} = \frac{3}{2}ln(x^{\frac{2}{3}}+1)$ .

EDIT: Oh well. Did your answer agree with mine?
March 3rd 2009, 12:57 AM
sleepiiee

yeah I got the same answer but I did it another way

Factores a x^(1/3) from the bottom and u subbed u = x^(2/3) + 1

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