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$\displaystyle \int \left( \frac{1}{x^2+3} \right)\; dx$

I've let $u=x^2+3$ but can't seem to get the right answer.

Really not sure what to do.

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Would you happen to remember what the derivative of the arctangent looked like? –  J. M. May 17 '11 at 16:47
    
I know it's similar, the 3 would be a 1. Alright, let's see if I can remember what chapter that was. –  eax May 17 '11 at 16:51
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3 Answers 3

up vote 4 down vote accepted

Hint: Note that $$\frac{1}{x^{2}+3}=\frac{1}{3}\cdot \frac{1}{\left( \frac{x}{\sqrt{% 3}}\right) ^{2}+1}$$

and use the substitution $u=\frac{x}{\sqrt{3}}$.

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Or please try putting $x = \sqrt{3} \tan\theta$. Then you have $dx = \sqrt{3}\cdot\sec^{2}\theta \ d\theta$. So you have your integral as:

\begin{align*} \int\frac{1}{x^{2}+3} \ \text{dx} &= \int \frac{\sqrt{3} \cdot \sec^{2}\theta}{3 (1+\tan^{2}\theta)} \ \text{d}\theta \\ &=\frac{1}{\sqrt{3}} \int \text{d}\theta \end{align*}

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I suggest letting $u = \frac{x}{\sqrt{3}}$. Then $du = \frac 1{\sqrt{3}} dx$ and $$ \int \left( \frac 1{x^2+3} \right) dx = \int \left( \frac 1{(\sqrt 3 u)^2 + 3} \, \sqrt 3 \right) dx = \frac 1{\sqrt{3}} \int \left( \frac 1{u^2 + 1} \right) = \frac 1{\sqrt{3}} \arctan(u) + C. $$ Hope that helps

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Thank you, but Americo's answer helped me understand how to choose u. –  eax May 17 '11 at 17:11
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