# Simple integral, can't get the right answer....

In summary, the conversation discusses the integral $\int \frac y {x^2+y^2} dx$ and how it is different from $\frac 1 y * \int \frac 1 {\frac {x^2}{y^2} + 1} dx = \frac 1 y * atan(x/y)$. It is explained that the latter is correct and the former is wrong because the variable of integration, x, is the independent variable and y is the dependent variable. However, the mentor in the forum suggests another way to write the integral in order to avoid confusion.f

Homework Statement
$$\int \frac y {x^2+y^2} dx$$
Relevant Equations
Just math
$$\int \frac y {x^2+y^2} dx$$
$$\frac 1 y * \int \frac 1 {\frac {x^2}{y^2} + 1} dx = \frac 1 y * atan(x/y)$$

The answer is just atan(x/y), which you get using u-substitution but I honestly don't see why I don't get it doing it the normal way.

It's because ##\displaystyle{\int} \dfrac{1}{\frac{x^2}{y^2} + 1} dx \neq \mathrm{arctan}\left( \dfrac{x}{y} \right)##. It would be correct to say, for example,\begin{align*}
\int \dfrac{1}{\frac{x^2}{y^2} + 1} dx = y \int \dfrac{1}{\frac{x^2}{y^2} + 1} d\left( \frac{x}{y} \right) = y\mathrm{arctan}\left( \frac{x}{y} \right)
\end{align*}

Last edited:
• • Delta2, FactChecker, Orodruin and 2 others
It was already in standard integral format. Try differentiating your answer and you'll see that's not right.

• FactChecker
I don't understand this thread. According to the OP
Homework Statement:: ##I = { \large \int \frac y {x^2+y^2} } dx~\dots~##
Since
##{~~~~}x = \left \{ \begin{align} & \text {variable of integration } \nonumber \\ &~~ \text {and is therefore the} \nonumber \\ & ~\text {independent variable} \nonumber \end{align} \right \}~~\Rightarrow~## ##\left \{ \begin{align} & y = y(x)~\text {because}~y~\text {is} \nonumber \\ & \text {normally used as the} \nonumber \\ & ~ \text {dependent variable} \nonumber \end{align} \right \}##
then
##{~~~~}I = { \large \int \frac y {x^2+y^2} } dx~\Rightarrow~I = { \large \int \frac {y(x)} {x^2+y^2} } dx##
so that integration by parts, wherein the right-hand side is a function of the independent variable only, no longer applies. But according to the forum mentor in post #2,
##\dots##. It would be correct to say, for example,\begin{align*}
\int \dfrac{1}{\frac{x^2}{y^2} + 1} dx = y \int \dfrac{1}{\frac{x^2}{y^2} + 1} d\left( \frac{x}{y} \right) = y\mathrm{arctan}\left( \frac{x}{y} \right)
\end{align*}
indicating that ##~y~## is actually a constant since
##{~~~~}d \left( { \large \frac {x}{y} } \right) = { \large \frac {dx}{y} }~\Rightarrow~y { \large \int } \dfrac{1}{\frac{x^2}{y^2} + 1} d\left( { \large \frac{x}{y} } \right) = { \large \int } \dfrac{1}{\frac{x^2}{y^2} + 1} dx##
If that is the case, then the original integral should have been written as
##{~~~~}I = { \large \int \frac k {x^2+k^2} } dx~\Leftarrow~k = \rm {constant}##
right at the start, or early on a remark posted by people who know more, in order to avoid confusion and waste of time.

• Delta2