Răspuns:
Integrala se mai scrie
[tex]I(x)=\displaystyle\int\dfrac{x^3dx}{x^4\sqrt{x^4-1}}[/tex]
[tex]\sqrt{x^4-1}=t\Rightarrow x^4-1=t^2\Rightarrow 4x^3dx=2tdt\Rightarrow x^3dx=\dfrac{1}{2}tdt[/tex]
[tex]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}[/tex]
Atunci
[tex]I(x)=\dfrac{1}{2}\arctan\sqrt{x^4-1}+\mathcal{C}[/tex]
Explicație pas cu pas:
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Răspuns:
Integrala se mai scrie
[tex]I(x)=\displaystyle\int\dfrac{x^3dx}{x^4\sqrt{x^4-1}}[/tex]
[tex]\sqrt{x^4-1}=t\Rightarrow x^4-1=t^2\Rightarrow 4x^3dx=2tdt\Rightarrow x^3dx=\dfrac{1}{2}tdt[/tex]
[tex]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}[/tex]
Atunci
[tex]I(x)=\dfrac{1}{2}\arctan\sqrt{x^4-1}+\mathcal{C}[/tex]
Explicație pas cu pas: