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fractions partielles ∫0132x2−64dx
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Solution
2(ln(7)−2ln(3))
+1
Decimal
−0.50262…
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Solve by:
One step at a time
∫0132x2−64dx
Extraire la constante: ∫a·f(x)dx=a·∫f(x)dx
=32·∫011x2−64dx
Prendre la fraction partielle de 1x2−64:−116(x+8)+116(x−8)
=32·∫01−116(x+8)+116(x−8)dx
Appliquer la règle de la somme: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx