Nei seguenti integrali è stata omessa la costante arbitraria C.

$$ \int x^\alpha \, \mathrm{d} x = \frac{x^{\alpha+1}}{\alpha + 1}$$

$$ \int \sqrt{ 1 + a \, x^2} \, \mathrm{d} x $$

integrali indefiniti funzioni radici calcolo

$$ a > 0 $$

Procediamo per sostituzione

$$ x = \frac{\tan u}{\sqrt{a} } \Longrightarrow \mathrm{d} x = \frac{\sec^2 u}{\sqrt{a}} \, \mathrm{d} u$$

Dalla relazione tra secante e tangente

$$ \sec^2 u = 1 + \tan^2 u $$

Sostituendo nell'integrale

$$ \int \sqrt{ 1 + a \, x^2} \, \mathrm{d} x = \int \sqrt{ 1 + \tan^2 u } \, \frac{\sec^2 u}{\sqrt{a}} \, \mathrm{d} u = \, \\ \, = \frac{1}{\sqrt{a}} \int \sec^3 u \, \mathrm{d} u = \frac{1}{\sqrt{a}} \left( \frac{\sin u \, \sec^2 u}{2} + \int sec^2 u \mathrm{d} u \right) = \, \\ \, = \frac{1}{2 \, \sqrt{a}} \left[ \tan u \, \sec u + \log \left( \tan u + \sec u \right) \right] = \, \\ \, = \frac{1}{2 \, \sqrt{a}} \left[ \tan u \, \sqrt{1 + \tan^2 u } + \log \left( \tan u + \sqrt{1 + \tan^2 u } \right) \right] $$

$$ \int \cos x \, \mathrm{d} x = \sin x$$

$$ \int \sin x \, \mathrm{d} x = - \cos x$$

$$ \int \cos^2 x \, \mathrm{d} x = \frac{x}{2} + \frac{\sin(2x)}{4}$$

$$ \int \sin^2 x \, \mathrm{d} x = \frac{x}{2} - \frac{\sin(2x)}{4}$$

$$ \int \sin x \cos x \, \mathrm{d} x = - \frac{cos(2x)}{4}$$

$$ \int \cos^3 x \, \mathrm{d} x = \sin x - \frac{\sin^3 x}{3}$$

$$ \int \sin^3 x \, \mathrm{d} x = - \cos x + \frac{\cos^3 x}{3}$$

$$ \int \cos x \, sin^2 x \, \mathrm{d} x = \frac{\sin^3 x}{3}$$

$$ \int \cos^2 x \, \sin x \, \mathrm{d} x = - \frac{\cos^3 x}{3}$$

$$ \int \cosh x \, \mathrm{d} x = \sinh x$$

$$ \int \sinh x \, \mathrm{d} x = \cosh x$$