Question

folosind schimbarea de variabila. Mulțumesc anticipat

Answer 1

Răspuns:

Integrala se mai scrie

$I(x)=\displaystyle\int\dfrac{x^3dx}{x^4\sqrt{x^4-1}}$

$\sqrt{x^4-1}=t\Rightarrow x^4-1=t^2\Rightarrow 4x^3dx=2tdt\Rightarrow x^3dx=\dfrac{1}{2}tdt$

$I(t)=\displaystyle\dfrac{1}{2}\int\dfrac{tdt}{(t^2+1)t}=\dfrac{1}{2}\int\dfrac{dt}{t^2+1}=\dfrac{1}{2}\arctan t+\mathcal{C}$

Atunci

$I(x)=\dfrac{1}{2}\arctan\sqrt{x^4-1}+\mathcal{C}$

Explicație pas cu pas: