Question

Resolva a integral:

$\Large\begin{array}{lr} \int \frac{dx}{\sqrt{x^2 -5x +6} } \end{arra}$


Só os ferras em :)

Answer 1

Conteúdo:

       Integração Por Substituição.

Resolução:

• Podemos substituir o numerador por 1 e deixar dx ao lado que seria:

$\large {\text { \sf \displaystyle \int\limits \cfrac{1}{\sqrt{x^2-5x+6} }\: \: dx }}$

• Observando a equação temos de primeiro completar o quadrado de $\large { \bf x^2-5x+6 }$

$\large { \sf \checkmark \quad \left( x- \cfrac{5}{2} \right )^2- \cfrac{1}{4} }$

• Que transforma em:

$\large {\text { \sf \displaystyle \int\limits \cfrac{1}{\sqrt{ \left ( x- \cfrac{5}{2} \right)^2 -\cfrac{1}{4} } }\:\:dx }}$

• Aplica integral de substituição:

$\large {\boxed { \sf u = x- \cfrac{5}{2} }}$

                     $\searrow$

                         $\large {\text { \sf \displaystyle \int\limits \cfrac{2}{\sqrt{4u^2-1}}\:\: du }}$

• Remover a constante:

$\large {\boxed { \sf \displaystyle \int\limits a\cdot f\left(x\right)dx=a\cdot \int\limits f\left(x\right)dx }}$

                   $\downarrow$

                 $\large {\text { \sf 2\cdot \displaystyle \int\limits \cfrac{1}{\sqrt{4u^2-1}} \:\: du }}$

• Temos de aplicar a integral de substituição:

       

$\large {\boxed { \sf u = \cfrac{1}{2}\: sec(v) }}$

                   $\large { \sf 2\cdot \displaystyle \int\limits \cfrac{sec \left(v\right)}{2}dv }$

• Remover a constante:

$\large { \sf 2\cdot \cfrac{1}{2}\cdot \displaystyle \int sec \left(v\right)dv }$

• Aplica as regras de integração:

$\large {\boxed { \sf \displaystyle \int sec \left(v\right)dv= ln \left \mid tan \left(v\right)+ sec \left(v\right)\right \mid }}$

                   $\large { \sf 2\cdot \cfrac{1}{2} \: ln \left \mid tan \left(v\right)+ sec \left(v\right)\right \mid }$

• Substitui na equação:

$\large { \sf 2 \cdot \cfrac{1}{2} \: ln \left| tan \left(arcsec \left(2\left(x-\cfrac{5}{2}\right)\right)\right)+ sec \left(arcsec \left(2\left(x-\cfrac{5}{2}\right)\right)\right ) \right | }$

• Simplifica:

$\large { \sf ln \left \biggm| 2\sqrt{x^2-5x+6}+2x-5\right | }$

• Para finalizar, adicionamos uma constante:

$\large {\boxed { \boxed{ \bf ln \left \biggm |2\sqrt{x^2-5x+6}+2x-5\right \biggm|+ \:C }}}$

Answer 2

$\large\boxed{\begin{array}{l}\underline{\rm completando~os~quadrados~de~ax^2+bx+c}\\\sf completando~os~quadrados~da~express\tilde ao\\\sf ~ax^2+bx+c\\\sf pode~ser~escrita~como~a\cdot\bigg(x+\dfrac{b}{2a}\bigg)^2-\dfrac{b^2-4ac}{4a}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm integrais~envolvendo~\sqrt{x^2-a^2}}\\\sf quando~o~integrando~\acute e~da~forma~\sqrt{x^2-a^2}\\\sf~use~a~substituic_{\!\!,}\tilde ao~x=a\,sec(\theta)\\\sf de~modo~que~dx=a\,sec(\theta)\cdot tg(\theta)\\\sf e~o~radicando~\sqrt{x^2-a^2}~se~torna~a\,tg(\theta).\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dx}{\sqrt{x^2-5x+6}}\\\underline{\rm completando~os~quadrados~temos}\\\sf x^2-5x+6=\bigg(x-\dfrac{5}{2}\bigg)^2-\dfrac{1}{4}\\\underline{\rm fac_{\!\!,}a}\\\sf t=x-\dfrac{5}{2}\\\sf dt=dx\\\displaystyle\sf\int\dfrac{dx}{\sqrt{x^2-5x+6}}=\int\dfrac{dt}{\sqrt{t^2-\dfrac{1}{4}}}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm fac_{\!\!,}a}\\\sf t=\dfrac{1}{2}\,sec(\theta)\\\sf dt=\dfrac{1}{2}\,sec(\theta)\cdot tg(\theta)\,d\theta\\\sf \sqrt{t^2-\dfrac{1}{4}}=\dfrac{1}{2}\,tg(\theta)\\\displaystyle\sf\int\dfrac{dt}{\sqrt{t^2-\dfrac{1}{4}}}=\int\dfrac{\backslash\!\!\!\frac{1}{2}\,sec(\theta)\cdot \diagup\!\!\!\!\!\!tg(\theta)}{\backslash\!\!\!\frac{1}{2}\,\diagup\!\!\!\!\!tg(\theta)}~d\theta\\\displaystyle\sf\int sec(\theta)~d\theta=\ell n|sec(\theta)+tg(\theta)|+k\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm usando~o~tri\hat angulo~auxiliar~temos:}\\\\\sf sec(\theta)=\dfrac{t}{\frac{1}{2}}=2t=2\cdot\bigg(x-\dfrac{5}{2}\bigg)=2x-5\\\\\sf tg(\theta)=\dfrac{\sqrt{t^2-\dfrac{1}{4}}}{\frac{1}{2}}=2\sqrt{t^2-\dfrac{1}{4}}\\\sf =2\sqrt{x^2-5x+6}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm isto~\acute e}\\\displaystyle\sf\int sec(\theta)d\theta=\ell n\bigg|2t+2\sqrt{t^2-\dfrac{1}{4}}\bigg|+k\\\underline{\rm Por~fim:}\\\displaystyle\sf\int\dfrac{dx}{\sqrt{x^2-5x+6}}=\ell n\bigg|2x-5+2\sqrt{x^2-5x+6}\bigg|+k\end{array}}$