Question

integrate step by step (x/(x^2 + y^2) ) dy

Answer 1

since it's dy then x and x^2 are just constants
\[x \int \frac{1}{x^2 + y^2} dy\]
isnt that solveable by one of the arc trig functions now?

Answer 2

I seen the answer to this problem on wolframalpha but have trouble getting there

Answer 3

i tried trig sub on this, may be i'm subing the wrong identiy in

Answer 4

\[\tan \theta = \frac yx\]
\[x \tan \theta = y\]
\[x\sec^2 \theta d\theta = dy\]

\[\sec \theta = \frac{\sqrt{x^2 + y^2}}{x}\]
\[x \sec \theta = \sqrt{x^2 + y^2}\]
\[x^2 \sec^2 \theta = x^2 + y^2\]

so your integral becomes
\[x \int \frac{x\sec^2 \theta d\theta}{x^2 \sec^2 \theta}\]
\[x \int \frac{d \theta }{x}\]
\[\implies \int d\theta\]
\[\implies \theta\]
\[\implies \tan^{-1}\left( \frac yx \right)\]
is that what wolfram got?