How to express in close form the parametric integral $\int{\frac{dy}{(x^2+y^2)^\frac{3}{2}}}$

Question

I met in Physics the following integral:

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

Does the following looks a valid approach?

Answer 1

Your answer is perfectly correct. Refining it further, notice that when $$\tan \theta = \frac{y} {x}$$ we have $$\sin \theta = \frac{y} {\sqrt{y^2+x^2}}$$ Hence, the integral equals: $$\boxed{I = \frac{y} {x^2\sqrt{x^2+y^2}}+C} $$

Answer 2

Well, we have:

$$\mathcal{I}_{\space\text{n}}:=\int\frac{1}{\left(\text{n}+x^2\right)^\frac{3}{2}}\space\text{d}x\tag1$$

Now, let's substitute $\text{u}:=\arctan\left(\frac{x}{\sqrt{\text{n}}}\right)$:

$$\mathcal{I}_{\space\text{n}}=\frac{1}{\text{n}}\cdot\int\cos\left(\text{u}\right)\space\text{d}\text{u}=\frac{\sin\left(\text{u}\right)}{\text{n}}+\text{C}=$$ $$\frac{\sin\left(\arctan\left(\frac{x}{\sqrt{\text{n}}}\right)\right)}{\text{n}}+\text{C}=\frac{x}{\sqrt{\text{n}}}\cdot\frac{1}{\sqrt{1+\left(\frac{x}{\sqrt{\text{n}}}\right)^2}}+\text{C}\tag1$$

Answer 3

You can find the result by computing only derivatives: $$ \begin{align} \int{dy\over(x^2+y^2)^{3/2}}=&-{1\over x}{d\over dx}\int{dy\over\sqrt{x^2+y^2}}\\ =&-{1\over x}{d\over dx}\int{d(y/x)\over\sqrt{1+(y/x)^2}}\\ =&-{1\over x}{d(y/x)\over dx}{d\over d(y/x)}\int{d(y/x)\over\sqrt{1+(y/x)^2}}\\ =&-{1\over x}{d(y/x)\over dx}{1\over\sqrt{1+(y/x)^2}}\\ =&-{1\over x}{-y\over x^2}{1\over\sqrt{1+(y/x)^2}}\\ =&{y\over x^2}{1\over\sqrt{x^2+y^2}}.\\ \end{align} $$