1. ## Indefinite integral

Can anyone help me with this integral?

$\displaystyle \int\frac{\1}{x^2-3}dx$

Thanks!

2. ## Re: Indefinite integral

$\displaystyle \int\frac{1}{x^2-a^2}~dx=\frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$

You can generate a proof of this formula in reverse with the chain rule; I was presented with such a proof and then taught the standard form.

Thaaanks!

4. ## Re: Indefinite integral

You can use use partial fractions to prove that formula.

$\displaystyle \frac{1}{x^2-a^2}=\frac{1}{(x-a)(x+a)}$

$\displaystyle \frac{1}{(x-a)(x+a)} \equiv \frac{A}{x-a}+\frac{B}{x+a}$

$\displaystyle \implies 1 \equiv A(x+a)+B(x-a)$

Putting x= a, -a we obtain

$\displaystyle A=\frac{1}{2a}$ and $\displaystyle B=-\frac{1}{2a}$

$\displaystyle \frac{1}{(x-a)(x+a)} = \frac{1}{2a}\Big[ \frac{1}{x-a}-\frac{1}{x+a} \Big]$

\displaystyle \begin{align*} \int \frac{1}{x^2-a^2}dx &=\int \frac{1}{(x-a)(x+a)}dx=\frac{1}{2a}\int \left( \frac{1}{x-a}-\frac{1}{x+a} \right)dx \\ &= \frac{1}{2a} (\ln|{x-a}|-\ln{|x+a|})+C \\ &= \frac{1}{2a} \ln{\Big| \frac{x-a}{x+a} \Big|}+C \end{align*}

5. ## Re: Indefinite integral

It's perfect!! Thanks!

6. ## Re: Indefinite integral

I have a little problem: I did the same that you said with this integral (image), but when I checked it with Wolfram, there's a -1 which I don't know where come from.

7. ## Re: Indefinite integral

What do you get if you do the long division?

8. ## Re: Indefinite integral

The same but without $\displaystyle -x$:$\displaystyle 6ln|x+2|-2ln|x+a|$

9. ## Re: Indefinite integral

Originally Posted by gulamerasool
The same but without $\displaystyle -x$:$\displaystyle 6ln|x+2|-2ln|x+a|$
\displaystyle \begin{align*} \frac{x-x^2}{(x+1)(x+2)} &= -\frac{(x^2+3x+2)-(4x+2)}{x^2+3x+2} \\ &= -1+\frac{4x+2}{(x+1)(x+2)} \\ &= -1+ \frac{4(x+1)-2}{(x+1)(x+2)} \\ &= -1+\frac{4}{x+2}-\frac{2}{(x+1)(x+2)} \\ &= -1+\frac{6}{x+2}-\frac{2}{x+2}\end{align*}

Integrate each term separately and see what you will get.