$u=\ln x;\ \ \ \ \ \ \ \ \ \ dv=xdx$

$u=xdx;\ \ \ \ \ \ \ dv=\ln xdx$

$\int_{ }^{ }x\ln xdx=\left(\ln x\right)\left(\frac{x^2}{2}\right)-\int_{ }^{ }\left(\frac{x^2}{2}\right)\left(\frac{dx}{x}\right)=$

$\int_{ }^{ }x\ln xdx=\left(x\right)-\int_{ }^{ }\ln xdx$

$\int_{ }^{ }x\ln xdx=\frac{x^2\ln x}{2}-\frac{1}{2}\int_{ }^{ }xdx=$

$\int_{ }^{ }x\ln xdx=\left(x\right)-x\ln x-\int_{ }^{ }dx$

$\int_{ }^{ }x\ln xdx=\frac{x^2\ln x}{2}-\frac{1}{2}\left(\frac{x^2}{2}\right)+c$

$\int_{ }^{ }x\ln xdx=\left(x\right)-x\ln x-x+c$

 $u^2=x^2;\ \ \ \ \ \ u=x;\ \ \ du=dx\ \ \ \ a^2=9;\ \ a=3$  

 $u^2=9;\ u=3;\ a^2=x^2;\ a=x;\ \ du=dx$  

 $\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}du}{u^2}$  

 $\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{u^2-a^2}du}{a^2}$  

 $\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}du}{u^2}\ caso\ I$ 

 $\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{u^2-a^2}du}{a^2}\ caso\ III$ 

$u=asenz;\ \ \ \ du=a\cos zdz;\ \ \ \sqrt{ }=a\cos z$

$u=a\sec z;\ \ \ \ du=a\sec ztgzdz:\ \ \ \ \ \sqrt{ }=atgz$

 $\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}dz}{u^2}=\int_{ }^{ }\frac{a\cos za\cos zdz}{a^2sen^2z}$ 

 $\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}}{u^2}=\int_{ }^{ }\frac{atgza\sec ztgzdz}{a^2\sec^2z}$ 

$\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}dz}{u^2}=\int_{ }^{ }\operatorname{ctg}^2zdz$

$\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}}{u^2}=\int_{ }^{ }\frac{tg^2zdz}{\sec^2z}$

$\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}dz}{u^2}=\int_{ }^{ }\csc^2zdz-\int_{ }^{ }dz$

$\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}}{u^2}=\int_{ }^{ }dz-\int_{ }^{ }\cos^2zdz$

 $1+tg^2z$  

 $tg^2z-1$  

 $\sec^2z-1$  

 $\sec^2z+1$  

 $1-\cos^2z$  

 $\cos^2z+1$  

 $1-sen^2z$  

 $sen^2z+1$  

 $1+\operatorname{ctg}^2z$  

 $\cot^2z-1$  

 $\csc^2z-1$  

 $\csc^2z+1$  

$\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}dz}{u^2}=-\operatorname{ctg}z-z+c$

$\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\int_{ }^{ }\frac{\sqrt{a^2-u^2}}{u^2}=z+\frac{1}{2}senz\cos z-\frac{1}{4}\cos z+c$

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$\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=-\frac{\sqrt{a^2-u^2}}{u}-arcsen\ \frac{u}{a}+c$

$\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\operatorname{arcsec}\ \frac{u}{a}-\frac{1}{2}\left(\frac{\sqrt{u^2-a^2}}{a}\right)\left(\frac{a}{u}\right)+c$

$\int_{ }^{ }\frac{\sqrt{9-x^2}dx}{x^2}=-\frac{\sqrt{9-x^2}}{x}-arcsen\ \frac{x}{3}+c$

$\int_{ }^{ }\frac{\sqrt{9-x^2}}{x^2}=\operatorname{arcsec}\ \frac{3}{x}-\frac{1}{2}\left(\frac{\sqrt{9-x^2}}{x}\right)\left(\frac{x}{3}\right)+c$

 $u=\log x;\ \ \ \ dv=dx$  

 $u=dx;\ \ \ \ \ \ \ dv=\log xdx$  

 $u^2=x^2;\ \ \ \ \ a^2=49$  

 $u^2=49;\ \ \ \ \ a^2=x^2$  

 $\int_{ }^{ }\sqrt{u^2+a^2}du=$  

 $\int_{ }^{ }\sqrt{a^2-u^2}dx=$