# [SOLVED] Integral and Error Bounds

• Mar 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
• Mar 3rd 2009, 12:04 AM
sleepiiee
• Mar 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, $\int\frac{dx}{x+x^{\frac{1}{3}}} = \frac{3}{2}ln(x^{\frac{2}{3}}+1)$.