Answer to Question #286832 in Calculus for Moon

Question

$\intop$ ²_-1 1/ 3√x2 dx=?

Answer 1

${\int_{-1}^{2} {\frac 1 {3\sqrt{x^2}}} dx}={\int_{-1}^{2} {\frac 1 {3*|x|}} dx}={\int_{-1}^{0} {\frac 1 {3*(-x)}} dx}+{\int_{0}^{2} {\frac 1 {3*x}} dx}=-{\frac 1 3}{\int_{-1}^{0} {\frac 1 x} dx}+{\frac 1 3}{\int_{0}^{2} {\frac 1 x} dx}=-{\frac 1 3}*ln|x||_{x=-1}^{x=0}+{\frac 1 3}*ln|x||_{x=0}^{x=2}=-{\frac 1 3}*{\lim_{x\to{0^-}} ln(-x)}+{\frac 1 3}*ln(1)+{\frac 1 3}*ln(2)-{\frac 1 3}*{\lim_{x\to{0^+}} ln(x)}$

Since ${\lim_{x\to0^+} lnx}=-\infty$, then this integral is divergent and equals $\infty$