I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$
I have been looking for a possible solution but they are with trigonometric integration..
I need a solution for this function without trigonometric integration
$\int\frac{\mathrm{d}x}{{(4+x^2)}^{3/2}}$
Let $I=\int \frac{dx}{\sqrt{x^2+4}} \,;\, J=\int \frac{dx}{\sqrt{x^2+4}^3}$
We integrate $I$ by parts, we get:
$u=(x^2+4)^{-1/2}, dv=dx$
$du=-x(x^2+4)^{-3/2}, v=x$
Thus
$I= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2}{\sqrt{x^2+4}^3}= \frac{x}{\sqrt{x^2+4}}+ \int \frac{x^2+4-4}{\sqrt{x^2+4}^3}=$ $= \frac{x}{\sqrt{x^2+4}}+ I-4J=$
Thus, canceling the $I$ we get
$4J= \frac{x}{\sqrt{x^2+4}} +C$
$\int\frac{dx}{(a^2+x^2)^{\frac n2}}=\int1\cdot \frac1{(a^2+x^2)^{\frac n2}}dx$
$=\frac x{(a^2+x^2)^{\frac n2}}-\int\left(\frac{-n}2\frac{2x\cdot x}{(a^2+x^2)^{\frac n2+1}} \right) dx$
$=\frac x{(a^2+x^2)^{\frac n2}}+n\int \left(\frac{(a^2+x^2-a^2)}{(a^2+x^2)^{\frac n2+1}}\right)dx $
$=\frac x{(a^2+x^2)^{\frac n2}}+n\left(\int\frac{dx}{(a^2+x^2)^{\frac n2}}-a^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}\right)+c $ where $c$ is the constant for indefinite integration.
or, $na^2\int\frac{dx}{(a^2+x^2)^{\frac n2+1}}=\frac x{(a^2+x^2)^{\frac n2}}+(n-1)\int\frac{dx}{(a^2+x^2)^{\frac n2}}+c$
Putting $n=1,$ $a^2\int\frac{dx}{(a^2+x^2)^{\frac 32}}=\frac x{(a^2+x^2)^{\frac 12}}+c$
$\frac{1}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\cdot\frac{1}{\left(1+\left(\frac{x}{2}\right)^2\right)^{3/2}}$
Now try
$x=2\sinh u\implies dx=2 \cosh u\,du\implies$
$\int\frac{dx}{\left(4+x^2\right)^{3/2}}=\frac{1}{8}\int\frac{2\,du\cosh u}{(1+\sinh^2u)^{3/2}}=\frac{1}{4}\int\frac{du}{\cosh^2u}=\ldots $
The following is not quite right, it needs minor modification for negative $x$. Let $x=\dfrac{1}{t}$. Then $dx=-\dfrac{1}{t^2}$. Substitute and do some algebra. There is some nice cancellation, and we end up with $\int \frac{-t\,dt}{(4t^2+1)^{3/2}}.$ Now let $u=4t^2+1$.