I check my answer but i found myself wrong. Here is the problem:
$$\int_0^ {1.8} \frac{1}{\sqrt{x}(1+x)}dx$$.
substitute $ u = \sqrt{x} ,\frac{dx}{\sqrt{x}}= 2du$
$$\int_0^{1.8} \frac{2du}{(1+u^2)}du$$
using arc tangent subs
$2 \arctan (\sqrt{x})$ then i plug the suitable number and ignore zero since, $ \arctan (0) = 0 $. and here what's happen when i check using calculator , it's may seems trivial but this is supposed definite integral , so can someone explain whats wrong here?
On
Note that the simple integral $$I=\int_{0}^{1} \frac{1}{1+x^2}dx=\pi/4$$ The value of $\pi/4=0.7853981633974483...$, whereas mathematica's NIntegrate gives a value of 0.785398163397447... which matches with the most accurate (analytic) result of $\pi/4$ but only up to 14 places after decimal. Numerical methods are supposed to be only approximate. One method can be more of less accurate than another.
Some numerical methods are Trapezoidal Rule, Riemann sum and Gaussian quadrature. See
My hypothesis for this is that Desmos as a graphing calculator uses a very high-accuracy numerical method for calculating the integral, and so their value for the integral is very slightly off.
Your $2\tan^{-1}{(1.8)}$ is definitely correct.