91影视

The integral is as follows.

${鈭�}_1^2鈥�\frac{1}{x\sqrt{9鈭�x^2}}dx$

The objective is to solve the integral.

To solve the integral, let $x=3\sin \left(u\right)$

Now, derivation of the above equation is solved below

$$\begin{gathered} x=3\sin \left(u\right) \\ dx=3\cos \left(u\right)du \end{gathered}$$

Now, use the identity $-{\sin }^2\left(u\right)+1={\cos }^2\left(u\right)$

So,

$\begin{array}{r}\frac{1}{x\sqrt{9鈭�x^2}}=\frac{1}{3\sqrt{鈭�9{\sin }^2鈦�\left(u\right)+9}\sin 鈦�\left(u\right)}\\ =\frac{1}{9\sqrt{鈭�{\sin }^2鈦�\left(u\right)+1}\sin 鈦�\left(u\right)}\\ =\frac{1}{9\sqrt{{\cos }^2鈦�\left(u\right)}\sin 鈦�\left(u\right)}\\ =\frac{1}{9\cos 鈦�\left(u\right)\sin 鈦�\left(u\right)}\end{array}$

The integral is solved below.

$\begin{array}{r}\frac{1}{3}鈭�\frac{1}{\sin 鈦�\left(u\right)}du\\ =\frac{1}{3}鈭�\left(\frac{1}{2\sin 鈦�\left(\frac{u}{2}\right)\cos 鈦�\left(\frac{u}{2}\right)}\right)du\end{array}$

$=\frac{1}{3}鈭�\left(\frac{1}{2\sin 鈦�\left(v\right)\cos 鈦�\left(v\right)}\right)dv\left[v=\frac{u}{2},dv=\frac{du}{2}\right]$

$=\frac{1}{3}鈭�\frac{{\sec }^2鈦�\left(v\right)}{\tan 鈦�\left(v\right)}dv$

$=\frac{1}{3}鈭�\frac{1}{w}dw\left[w=\tan 鈦�v,dw={\sec }^2鈦�\left(v\right)dv\right]$

$\begin{array}{r}=\frac{1}{3}\ln 鈦�\left(\right|w\left|\right)\\ =\frac{1}{3}\ln 鈦�\left(\right|\tan 鈦�\left(v\right)\left|\right)=\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{x}{3}\right)\right)\right|\right)\end{array}$

$=\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{u}{2}\right)\right|\right)$

The definite integral is solved below.

${鈭�}_1^2鈥�\frac{1}{x\sqrt{9鈭�x^2}}dx$

$={\left[\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{x}{3}\right)\right)\right|\right)\right]}_1^2$

$=\left[\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{2}{3}\right)\right)\right|\right)\right]鈭�\left[\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{1}{3}\right)\right)\right|\right)\right]$

Therefore, the value is$\left[\left[\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{2}{3}\right)\right)\right|\right]鈭�\left[\frac{1}{3}\ln 鈦�\left(\left|\tan 鈦�\left(\frac{1}{2}{\sin }^{鈭�1}鈦�\left(\frac{1}{3}\right)\right)\right|\right)\right]\right.\right.\text{.}$