Integrating Rational Functions with Trigonometric Substitutions

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How to integrate integral
\int\frac{8}{x^2+4}dx
?
 
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Rewrite it in the form:
\int\frac{2dx}{(\frac{x}{2})^{2}+1}
See if that helps you.
 
The derivative of arctan(x) is
\frac{d tan^{-1}(x)}{dx}= \frac{1}{x^2+ 1}

Does that help?
 
\int\frac{8}{x^2+4}dx=\int\frac{2dx}{(\frac{x}{2})^2+1}=\int\frac{4d(\frac{x}{2})}{(\frac{x}{2})^2+1}=4\arctan\frac{x}{2}
d(\frac{x}{2})=\frac{1}{2}dx
dx=2d(\frac{x}{2})
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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