91Ó°ÊÓ

An integral is a number that is assigned to a function in order to explain displacement, area, volume, and other concepts that arise when infinitesimal data is combined.

Write the expression for the integral.

$s={∫}_0^1\sqrt{1+4x^2}dx$

It is known that localid="1658831726900" $x=\frac{1}{2}\sin hu,dx=\frac{1}{2}\cos hudu$. Substitute these values in the above expression.

$$\begin{gathered} s={∫}_0^{×{\sin }^{-1}2}\sqrt{1+\sin h^{2}u}\frac{1}{2}\cos hudu \\ =\frac{1}{2}{∫}_0^{×{\sin }^{-1}2}\cos h^{2}udu \\ =\frac{1}{4}{∫}_0^{×{\sin }^{-1}2}\cos h\left(2u\right)+1)du \\ =\frac{1}{4}{\overline{)\left[u+\frac{1}{2}\sin h\left(2u\right)\right]}}_0^{\sin h^{-1}2} \\ =\frac{1}{4}\left[\sin h^{-1}2+\frac{1}{2}\sin h\left(2\sin h^{-1}2\right)\right] \end{gathered}$$

Write the expression for the identity.

$sin{h}^{-1}x=ln\left(x+\sqrt{x^2+1}\right)$

To get the following expression.

$$\begin{gathered} sin{h}^{-1}2=\ln \left(2+\sqrt{5}\right) \\ \sin h\left(2sin{h}^{-1}2\right)=\frac{1}{2}\left(ci{n}^{2\ln \left(2-\sqrt{5}\right)}-e^{-2\ln \left(2-\sqrt{5}\right)}\right) \\ =\frac{1}{2}\left[{\left(2+\sqrt{5}\right)}^2-\frac{1}{{\left(2+\sqrt{5}\right)}^2}\right] \\ =\frac{1}{2}\left[9+4\sqrt{5}-\frac{1}{9+4\sqrt{5}}\right] \\ =\frac{80+36\sqrt{5}}{9+4\sqrt{5}} \end{gathered}$$

As a result, equation (i) equals to the expression.

$\left(1\right)=\frac{1}{4}\left(\ln \left(2+\sqrt{5}\right)+\frac{40+18\sqrt{5}}{9+4\sqrt{5}}\right)$

But it is known, $\left(40+18\sqrt{5}\right)=2\sqrt{5}\left(9+4\sqrt{5}\right)$.

As a result, equation (i) equals to the expression.

$s=\frac{1}{4}\left(\ln \right(2+\sqrt{5})+2\sqrt{5}$

Thus, the provided problem's final solution is $s=\frac{1}{4}\left(\ln \right(2+\sqrt{5})+2\sqrt{5}$.