integral from 1 to 4 of (ax+1/x)^2

解答

\int_{1}^{4} (a x + \frac{1}{x})^{2} d x

a^{2} \cdot 21 + 6 a + \frac{3}{4}

求解步骤

\int_{1}^{4} (a x + \frac{1}{x})^{2} d x

化简

(a x + \frac{1}{x})^{2} : a^{2} x^{2} + 2 a + \frac{1}{x ^{2}}

= \int_{1}^{4} a^{2} x^{2} + 2 a + \frac{1}{x ^{2}} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{4} a^{2} x^{2} d x + \int_{1}^{4} 2 a d x + \int_{1}^{4} \frac{1}{x ^{2}} d x

\int_{1}^{4} a^{2} x^{2} d x = a^{2} \cdot 21

\int_{1}^{4} 2 a d x = 6 a

\int_{1}^{4} \frac{1}{x ^{2}} d x = \frac{3}{4}

= a^{2} \cdot 21 + 6 a + \frac{3}{4}

\int_{1}^{2} (\frac{x e ^{x}}{y}) d y

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