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- #1

#### rsoy

##### New member

- Mar 3, 2013

- 14

- Thread starter rsoy
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- Thread starter
- #1

- Mar 3, 2013

- 14

- Jan 17, 2013

- 1,667

the same idea here as well ,but i dont understand what you are doing exactly !

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- #3

- Mar 5, 2012

- 9,591

Welcome to MHB, rsoy!

Your problem seems to be that you assume $\dfrac{x+y}{x^2+y^2} = \dfrac 1 {x+y}$.

But this is not true.

So instead your next step for the first part of the expression would be (edited):

$$\begin{aligned} \int^y \frac{x+y}{x^2+y^2}dy &= \int^y \frac{x}{x^2+y^2}dy &&+ \int^y \frac{y}{x^2+y^2}dy &\\

&= \arctan \left(\frac y x \right) &&+ \frac 1 2 \ln(x^2+y^2) &+ C(x) \end{aligned}$$

Last edited:

- Jan 17, 2013

- 1,667

Shouldn't the resultant constant be a function of x !Welcome to MHB, rsoy!

Your problem seems to be that you assume $\dfrac{x+y}{x^2+y^2} = \dfrac 1 {x+y}$.

But this is not true.

So instead your next step for the first part of the expression would be:

$$\begin{aligned} \int^y \frac{x+y}{x^2+y^2}dy &= \int^y \frac{x}{x^2+y^2}dy &&+ \int^y \frac{y}{x^2+y^2}dy &\\

&= \arctan \left(\frac y x \right) &&+ \frac 1 2 \ln(x^2+y^2) &+ C \end{aligned}$$

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- #5

- Mar 5, 2012

- 9,591

Good point!Shouldn't the resultant constant be a function of x !

Edited.

- Thread starter
- #6

- Mar 3, 2013

- 14

thaaanks