Question

integral fraction numerator dx over denominator x square root of x to the power of 4 minus 16 end root end fraction

Answer 1

$\int\frac{dx}{x\sqrt{x^4-16}}\\ put \space x^4-16=t^2\\ 4x^3dx=2tdt\\ \int\frac{dx}{x\sqrt{x^4-16}}\\ =\frac{1}{2}\int\frac{dt}{t^2+16}\\ =\frac{1}{32}\int\frac{dt}{(\frac{t}{4})^2+1}\\ =\frac{4}{32}tan^{-1}(\frac{t}{4})+c\\ =\frac{1}{8}tan^{-1}(\frac{\sqrt{x^4-16} }{4})+c\\$