∫x√(x² + 1) dx
u = x² + 1
du/2 = x dx
1/2*∫√u du
1/3*(x² + 1)^(3/2) + C
Let u=x^2+1.
Then,
Integrate[x*Sqrt[x^2 + 1], x] = 1/2 Integrate[Sqrt[u], u] = u^(3/2)/3 = 1/3 (1 + x^2)^(3/2)
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∫x√(x² + 1) dx
u = x² + 1
du/2 = x dx
1/2*∫√u du
1/3*(x² + 1)^(3/2) + C
Let u=x^2+1.
Then,
Integrate[x*Sqrt[x^2 + 1], x] = 1/2 Integrate[Sqrt[u], u] = u^(3/2)/3 = 1/3 (1 + x^2)^(3/2)