Problem with indefinite integral

**lenric** · May 25th 2012, 08:27 PM

Hi guys!

I've got a problem with some indefinite integrals...
I don't want to solve them using neither the substitution method, nor the parts one, and that's why I'm experiencing some problems...

So for example, the integral of x/(1+4x^2) = 1/8 ln (4x^2+1), I get it, because (4x^2)=8x... however, I don't get why the integral of 1/(4x^2+1) is 1/2 arctan (2x), why is it 1/2 and why is it 2x?
The same doubt arises when it is let's say 1/(1+3x^2)...

I tried using the wolframalpha calculator, but it doesn't show the steps taken in order to achieve the right function...

I guess it has something to do with 1 being the numerator, instead of x, but even so I don't understand why...

Thank you guys!

**Pupil** · May 25th 2012, 09:25 PM

There's an integral formula:

$\frac{1}{a^2 + x^2} = \frac{1}{a}arctanx + C$

Where a is any arbitrary constant.

**Siron** · May 25th 2012, 10:15 PM

$\int \frac{1}{1+4x^2}dx$
$=\int \frac{1}{1+(2x)^2}dx$
Let $2x=t \Rightarrow 2dx=dt \Leftrightarrow dx=\fac{1}{2}dt$
thus the integral becomes
$\frac{1}{2} \frac{1}{1+t^2}dt = \frac{1}{2}\arctan(t)+C=\frac{1}{2}\arctan(2x)+C$

Originally Posted by Pupil

There's an integral formula:
$\frac{1}{a^2 + x^2} = \frac{1}{a}arctanx + C$
Where a is any arbitrary constant.

This not correct. It should be:
$\int \frac{dx}{a^2+x^2}=\int \frac{dx}{a^2\left(1+\frac{x^2}{a^2}\right)}=\int \frac{dx}{a^2\left[1+\left(\frac{x}{a}\right)^2\right]}$
Let $\frac{x}{a}=t \Rightarrow dx=adt$ thus the integral becomes
$\frac{a}{a^2} \int \frac{dt}{1+t^2}=\frac{1}{a}\arctan(t)+C=\frac{1}{ a}\arctan\left(\frac{x}{a}\right)+C$

**lenric** · May 26th 2012, 06:05 AM

Thank you guys!

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