微積分ヘルプ: 積分 ∫ 1/((x^2-1)^(3/2) ) dx - 三角関数の置換による積分

1. Integrales de por sustitución trigonométrica - ∫(1)/√(16-x^2)^(3/2) dx

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Calculus Help: Integral of ∫ (x^2 dx)/(x^2+16)^(3/2) - Integration by Trigonometric Substitution

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∫1/√(x² + x) dx = ? Solve without using substitution method.

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微積分ヘルプ: ∫√(16-x^2)dx の積分 - 三角関数の置換による積分

∫e^(-1/x)/x² dx [0, ∞] = ??

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統合 | 1/sqrt(16-x^2) dx

Calculus Help: Integral of (x^3 dx)/√(16-x^2 ) - Integration by substitution

1/sqrt(16-x^2) dx の積分

∫ ((36-25x^2 )^(3/2) )/x^6 dx - Integration by trigonometric substitution - 三角関数による置換による積分