l don't understand the d/dx in front of the integral... $\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$ = arctan 0 - arctan x = -arctanx could this be right?
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Originally Posted by algebra2 l don't understand the d/dx in front of the integral... $\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$ = arctan 0 - arctan x = -arctanx could this be right? You need to use the fundemental theorem of calculus $\displaystyle \frac{d}{dx} \int_{0}^{x}\frac{1}{1+u^2}du=\frac{1}{1+x^2}$
there was no reason to integrate and then differentiate, but you made mistakes in your integration.... Originally Posted by algebra2 l don't understand the d/dx in front of the integral... $\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$ = $\displaystyle {d\over dx} (\arctan x - \arctan 0)$ = $\displaystyle {d\over dx} (\arctan x) $ = $\displaystyle {1\over 1+x^2}$ is right.... .....NOW, but is unnecessary.
Last edited by matheagle; May 21st 2009 at 07:15 AM.
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