# improper integral

• October 25th 2009, 06:47 PM
jlmills5
improper integral
int 4/[sqrt(x)(x+6)] dx from 0 to infinity

0 and infinity are both problems...solution says to break it up as limits from 0 to 1 and 1 to infinity....why? how is this decided? Is a convenient number just decided upon?
• October 25th 2009, 07:06 PM
Krizalid
it can be any number as long as $x\in[0,\infty)$ and besides is usual and convenient.

we have $\int_{0}^{\infty }{\frac{dx}{(x+6)\sqrt{x}}}=\int_{0}^{1}{\frac{dx} {(x+6)\sqrt{x}}}+\int_{1}^{\infty }{\frac{dx}{(x+6)\sqrt{x}}},$ so in order to get convergence here, these pieces must converge.

the first one does by simple limit comparison test with $\int_0^1\frac{dx}{\sqrt x},$ and for the second one, for each $x\ge1$ is $\frac{1}{(x+6)\sqrt{x}}<\frac{1}{x\sqrt{x}}=\frac{ 1}{x^{3/2}},$ thus these pieces converge and the original integral does.