Best way to integrate?

**billym** · February 18th 2008, 08:46 AM

What would be the best approach to integrating:

x / (1 + x^2) , ( x > 0)

(is "x arctan x + C" a possible solution?)

By substituting u=(1+x^2) , is "ln sqrt (1 + x^2) + C" correct?

**Krizalid** · February 18th 2008, 08:51 AM

There's no way to get an arctangent there, it's just a hidden natural logarithm. Besides, your substitution works and your answer is correct.

**billym** · February 18th 2008, 08:56 AM

Thanks alot.

My problem then is:

Given:

ln sqrt (1+x^2) = y + 1/2(y^2) + C

how would I solve for y?

**Krizalid** · February 18th 2008, 09:00 AM

As far as I can see, you're probably solving a separable ODE.

If you want to isolate $y,$ calculus ends right here, it's just solving an algebra problem. You require to apply the quadratic formula. (Or may be completing the square could be helpful.)

**billym** · February 18th 2008, 09:25 AM

Sorry, I'm trying to isolate x, not y.

I have :

x^2 = (e^(y + 1/2(y^2) + C)^2) - 1

(let e^C = D)

x = sqrt ((D*e^y*e(1/2(y^2))^2 - 1))

Is this correct?

**mr fantastic** · February 18th 2008, 10:40 AM

Deleted - I misread the equation.

**mr fantastic** · February 18th 2008, 10:51 AM

Originally Posted by billym

Thanks alot.

My problem then is:

Given:

ln sqrt (1+x^2) = y + 1/2(y^2) + C

how would I solve for y?

Originally Posted by billym

Sorry, I'm trying to isolate x, not y.

I have :

x^2 = (e^(y + 1/2(y^2) + C)^2) - 1

Mr F says: $\sqrt{1 + x^2} = e^{y + \frac{y^2}{2} + C} \Rightarrow 1 + x^2 = (e^{y + \frac{y^2}{2} + C})^2 = e^{2 \left( y + \frac{y^2}{2} + C \right)} = e^{2y + y^2 + 2C}$

(let e^C = D)

Mr F says: No. Let $e^{2C} = D$ . Therefore $1 + x^2 = D e^{2y + y^2} \Rightarrow x^2 = D e^{2y + y^2} - 1$ .

And when you take the square root it'll be $x = \pm ........$ .

x = sqrt ((D*e^y*e(1/2(y^2))^2 - 1))

Is this correct?

Note: $(a^{m})^n = a^{mn}$ , NOT $a^{m^n}$ ....

**mr fantastic** · February 18th 2008, 05:58 PM

And in fact the answer can be simplified a bit:

$x^2 = D e^{2y + y^2} - 1 = D e^{(y + 1)^2 - 1} - 1 = D \, e^{-1} e^{(y+1)^2} - 1$ .

Let $D\, e^{-1} = A$ . Then:

$x^2 = A \, e^{(y+1)^2} - 1$ .

Therefore $x = \pm \sqrt{A \, e^{(y+1)^2} - 1}$ .

Alternatively, you can easily make y the subject if required .....

Thread: Best way to integrate?

LinkBack

Thread Tools

Display

Best way to integrate?

Similar Math Help Forum Discussions

Integrate 3 e^x

how to integrate this ??

Integrate

How to integrate x^x?

Integrate

Search Tags