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ถามโจทย์ปัญหา => ถามปัญหาคณิตศาสตร์ => Topic started by: Aguero on December 10, 2014, 09:15:34 AM



Title: Calculus ครับ
Post by: Aguero on December 10, 2014, 09:15:34 AM
 \displaystyle\int\sqrt{\frac{x+1}{x+3}} dx


Title: Re: Calculus ครับ
Post by: jali on December 10, 2014, 04:17:01 PM
\displaystyle \int \sqrt{\frac{x+1}{x+3}} dx
ให้ \displaystyle u=x+2
\displaystyle \int \sqrt{\frac{u-1}{u+1}} du
ให้ \displaystyle u=\sec^{2}(\theta)
\displaystyle \int \sqrt{\frac{\tan^{2}(\theta)}{2+\tan^{2}(\theta)}} 2 \sec(\theta)\tan(\theta)\sec(\theta) d\theta
\displaystyle \int \frac{2\tan^{2}(\theta)}{\sqrt{2+\tan^{2}(\theta)}} d\tan(\theta)
\displaystyle 2\left [ \int \sqrt{2+\tan^{2}(\theta)}d\tan(\theta)+2 \int \frac{1}{\sqrt{2+\tan^{2}(\theta)}} d\tan(\theta) \right ]
ให้ \displaystyle \tan(\theta)=\sqrt{2}\tan(\phi)
\displaystyle 2 \left[ \int 2 \sec^{3}(\phi) d\phi-2 \int \sec(\phi) d\phi \right]
\displaystyle \int \sec^{3}(\phi) d\phi=\frac{1}{2} \left( \sec(\phi)\tan(\phi) +\ \int \sec(\phi) d\phi  \right)
\displaystyle 2 \left ( \sec(\phi)\tan(\phi) -\ln(\abs{\sec(\phi)+\tan(\phi)}) \right )
\displaystyle 2 \left ( \frac{\tan(\theta)\sqrt{2+\tan^{2}(\theta)}}{2}-\ln(\abs{\frac{\tan(\theta)+\sqrt{2+\tan^{2}(\theta)}}{\sqrt{2}}} \right)
\displaystyle 2 \left ( \frac{\sqrt{x+1}\sqrt{x+3}}{2}-\ln(\frac{\sqrt{x+1}+\sqrt{x+3}}{\sqrt{2}} \right )