Equations are in $\LaTeX$ format; I'm still trying to understand how MathJax works.
Given the following integral:
$\int_0^{+\infty } \frac{1}{x \sqrt{x}} \, dx$
I'm pretty sure that does not converge.
And if the integral was a indefinite one, also I'm almost sure the result is $\ln(\sqrt x)$ using substitution.
If I am correct, why wolfram alpha says that the result is $\frac{-2}{\sqrt x}$ instead of $\ln(\sqrt x)$?
MathWay shows the correct results for both definite and indefinite integrals.
I'm new to any mathematical soft like Wolfram Alpha/Mathematica, and my idea is to test my pen and paper results with software that checks my results.
Any hints will be greatly appreciated.
Thanks.
Best Answer
If you are at the point of computing integrals, then you are probably already comfortable with computing derivatives. The way to check whether a potential antiderivative is correct is to take its derivative. If you take the derivative of $f(x)=\ln(\sqrt x)$ (either by first simplifying to $\frac{1}{2}\ln(x)$ or using the chain rule) you get $f'(x)=\frac{1}{2x}$. If you take the derivative of $g(x)=-\frac{2}{\sqrt x}=-2x^{-1/2}$, you get $g'(x)=x^{-3/2}=\frac{1}{\sqrt{x}^3}=\frac{1}{x\sqrt{x}}$.