integrate (7x^2 + 5)^-1

Question

abb0t · Answer

$\large\int\limits  \frac{1}{7x^2+5}dx$ is that correct?

anonymous · Answer

yup.

abb0t · Answer

play around with it to get it in the form of artan where you have $ax^2+1$ which means factor out the 5.

abb0t · Answer

can you try that first :) if you need help, let me know.

anonymous · Answer

OMG thanks :P

anonymous · Answer

but does then does it become 1/a tan^-1(x/a) or something?? Sorry, i've forgotten the general formula.

anonymous · Answer

@terenzreignz

abb0t · Answer

$$\int\limits \frac{1}{ax^2+1} = \arctan(x)$$

anonymous · Answer

I mean, what is the integral of $$\frac{ 1 }{ ax^2 + 1 }$$

anonymous · Answer

where does the a go??

abb0t · Answer

the "a" is the number in front of $x^2$!!

terenzreignz · Answer

Well, you could let $u = x\sqrt{a}$

And work from there.

anonymous · Answer

Hold on guys, the derivative of arctan is $$\frac{ 1 }{ x^2 + 1 }$$, so my question is that the a is a coefficient, but you need to multiply it or divide it to the arctan when you integrate it, so what do we do with it??

terenzreignz · Answer

Well...
$$\large \int \frac{dx}{ax^2+1}$$
$$let \ u = x\sqrt a\du = \sqrt a \ dx$$

$$\large = \frac1{\sqrt a}\int \frac{du}{u^2 +1}$$
And work from there.

anonymous · Answer

ohh. ok, lemme do that.

anonymous · Answer

|dw:1367417552288:dw|
This can be easily derived using the substitution$$x = a \tan \theta $$
Here 'a' is a constant... :-)