# Thread: [SOLVED] Integral and Error Bounds

1. ## [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

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

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

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

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

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

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

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

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

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