# indefinite integral with noninteger radical in denominator

I tried to use substitution and the power rule to no avail. How should I go about finding the integral of this?

${dy\over (x^2+y^2)^\frac 3 2 }$

The book I'm reading gives the answer but I want to know how to work it myself (not look it up in table).

${y\over x^2(x^2+y^2)^\frac 1 2 }$

Thank you, relayman357

SOLUTION:

Let y = x $\tan u$
u = $\arctan (y/x)$
$dy/du=x (sec u)^2$

$\int {x (sec^2 u) du\over [x^2+x^2 (tan^2 u)]^\frac 3 2 }$

$\int {x (sec^2 u) du\over x^3[1+tan^2 u]^\frac 3 2 }$

$\int {sec^2 u du\over x^2[sec^2u]^\frac 3 2 }$

$\int {sec^2 u du\over x^2[sec^3u]}$

$1/x^2\int {du\over sec u }$

$1/x^2\int {(cosu) du }$

$1/x^2 sin (u)$

Undoing the substitution and using the right-triangle relationships:
$1\over x^2 sin(arctan (y/x))$

$y\over x^2(x^2+y^2)^\frac 1 2$

• Have you tried the substitution $y=x\sinh(t)$, leading to $dy = x \cosh(t)\,dt$? Commented Apr 11, 2017 at 15:29
• No, I will see what I can do with your suggestion. By the way, why doesn't simple substitution work? e.g. u = ${(x^2+y^2)^\frac 3 2 }$ Commented Apr 11, 2017 at 16:57
• It turns out that if i had a "y" in the numerator i could have solved this with simple "u" substitution. e.g. ${y\over (x^2+y^2)^\frac 3 2 }$ can be easily solved with u = ${(x^2+y^2)}$. How do we know when that will work - are there any rules or guidelines? Commented Apr 13, 2017 at 14:35

Take $$x^2+y^2=y^2\left(\left(\frac{x}{y}\right)^2+1\right)$$ and substitute $$\frac{x}{y}=t$$