# Thread: Best way to integrate?

1. ## Best way to integrate?

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?

2. 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.

3. Thanks alot.

My problem then is:

Given:

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

how would I solve for y?

4. 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.)

5. 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?

6. Deleted - I misread the equation.

7. 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 $a^{m^n}$ ....

8. 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 .....