I need some help with integrating {1 dx / (x^2+3)^(3/2) }. I got
(1/3)[ln {(x^2+3)^(1/2) +x} - ln {(3)^(1/2)}] + C. I tried to differentiate this but could not get it back to original form.
$\displaystyle \int \frac{1}{(x^2+3)^{3/2}} \ dx$
I am going to do this in a strange way:
Instead find,
$\displaystyle \int \frac{1}{(x^2+1)^{3/2}} dx = \int \frac{1}{\sqrt{x^2+1}}\cdot \frac{1}{x^2+1} \ dx$
Use substitution, $\displaystyle t = \tan^{-1} x \Rightarrow t' = \frac{1}{x^2+1}$
With $\displaystyle \sec t = \frac{1}{\sqrt{x^2+1}}$
Thus,
$\displaystyle \int \sec t dt = \ln |\sec t + \tan t| + C$
Thus,
$\displaystyle \ln | \sec (\tan^{-1} x) + \tan (\tan^{-1} x) | + C = \ln | \sqrt{x^2+1} + x | + C$
Apparently this can also be done with inverse hyperbolic functions, but I think it is a waste to memorize them. So I did it the standard way.
Now given,
$\displaystyle \int \frac{1}{(x^2+3)^{3/2}} dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left( \frac{x^2}{3} + 1 \right)^{3/2}} \ dx = \frac{1}{3^{3/2}} \cdot \int \frac{1}{\left[ \left( \frac{x}{\sqrt{3}}\right)^2 + 1 \right]^{3/2}} \ dx$
Use the substitution $\displaystyle u= \frac{x}{\sqrt{3}}$ and above work to get:
$\displaystyle \frac{1}{3^{3/2}} \cdot \frac{\sqrt{3}}{1} \cdot \ln \left| \sqrt{ \frac{x^2}{3} + 1} + \frac{x}{\sqrt{3}} \right| + C = \frac{1}{3} \cdot \ln \left| \frac{\sqrt{x^2+3}}{\sqrt{3}} + \frac{x}{\sqrt{3}} \right|+C$
Use the properties of the natural logarithm,
$\displaystyle \frac{1}{3} \cdot \left( \ln | \sqrt{x^2+3} + x | - \ln \frac{1}{\sqrt{3}} \right)+C$
Combine constants,
$\displaystyle \frac{1}{3}\cdot \ln |\sqrt{x^2+3} + x | + C$