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CAL1 • Integrales
CAL1_INT_156
Guía de ejercicios - Tipo 3
Enunciado
Calcular:
$$ \int \frac{dx}{x^2(x^7 + 1)^{6/7}} $$
$$ \int \frac{dx}{x^2(x^7 + 1)^{6/7}} $$
Solución Paso a Paso
1. Desarrollo paso a paso:
Factorizamos $x^7$ del radical:
$$ (x^7 + 1)^{6/7} = [x^7(1 + x^{-7})]^{6/7} = x^6(1 + x^{-7})^{6/7} $$
Sustituyendo en la integral:
$$ \int \frac{dx}{x^2 \cdot x^6(1 + x^{-7})^{6/7}} = \int \frac{dx}{x^8(1 + x^{-7})^{6/7}} = \int x^{-8}(1 + x^{-7})^{-6/7} \, dx $$
Sea $u = 1 + x^{-7}$, entonces $du = -7x^{-8} \, dx$.
$$ \int u^{-6/7} \left( -\frac{du}{7} \right) = -\frac{1}{7} \cdot \frac{u^{1/7}}{1/7} + C = -u^{1/7} + C $$
2. Resultado final:
$$ \boxed{-\frac{\sqrt[7]{x^7 + 1}}{x} + C} $$
Factorizamos $x^7$ del radical:
$$ (x^7 + 1)^{6/7} = [x^7(1 + x^{-7})]^{6/7} = x^6(1 + x^{-7})^{6/7} $$
Sustituyendo en la integral:
$$ \int \frac{dx}{x^2 \cdot x^6(1 + x^{-7})^{6/7}} = \int \frac{dx}{x^8(1 + x^{-7})^{6/7}} = \int x^{-8}(1 + x^{-7})^{-6/7} \, dx $$
Sea $u = 1 + x^{-7}$, entonces $du = -7x^{-8} \, dx$.
$$ \int u^{-6/7} \left( -\frac{du}{7} \right) = -\frac{1}{7} \cdot \frac{u^{1/7}}{1/7} + C = -u^{1/7} + C $$
2. Resultado final:
$$ \boxed{-\frac{\sqrt[7]{x^7 + 1}}{x} + C} $$