I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is: Simplifying further, I get: and But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it.
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$\displaystyle \displaystyle \int \dfrac{1}{b^2+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C$
Last edited by harish21; Nov 14th 2010 at 07:59 AM. Reason: typo
Originally Posted by TwoPlusTwo I'm working on a pretty long integration problem, and I've managed to split it up into simpler parts. One of them is: Simplifying further, I get: and But I'm stuck on the last part. Wolfram says arctan something, but I need to know how do it. $\displaystyle \displaystyle{\int\frac{1}{2+x^2}\,dx=\frac{1}{\sq rt{2}}\int\frac{1/\sqrt{2}\,dx}{1+\left(\frac{x}{\sqrt{2}}\right)^2} }$ $\displaystyle \display{=\frac{1}{\sqrt{2}}\arctan\left(\frac{x}{ \sqrt{2}}\right)+C}$ Tonio
Originally Posted by harish21 $\displaystyle \displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{b} tan^{-1}\bigg(\frac{x}{b}\bigg)+C$ That holds if and only if $\displaystyle b = 1$. $\displaystyle \displaystyle \int \dfrac{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k$.
Originally Posted by TheCoffeeMachine That holds if and only if $\displaystyle b = 1$. $\displaystyle \displaystyle \int \dfrainc{1}{b+x^2} \;dx = \dfrac {1}{\sqrt{b}}\tan^{-1}\bigg(\frac{x}{\sqrt{b}}\bigg)+k$. indeed. it's usually written $\displaystyle \displaystyle \int \frac 1{b^2 + x^2}~dx = \frac 1b \tan ^{-1} \frac xb + C$ using $\displaystyle \displaystyle a$ instead of $\displaystyle \displaystyle b$
Correction already made by tonio [b]and[/b\] TheCoffeeMachine!
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