integral from 1 to infinity of 7/(x^2+4)

解答

\int_{1}^{\infty} \frac{7}{x ^{2} + 4} d x

\frac{7 π}{4} - \frac{7}{2} arctan (\frac{1}{2})

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3.87502 \dots

求解步骤

\int_{1}^{\infty} \frac{7}{x ^{2} + 4} d x

计算不定积分:

\int \frac{7}{x ^{2} + 4} d x = \frac{7}{2} arctan (\frac{x}{2}) + C

带入上下限计算后得:

\int_{1}^{\infty} \frac{7}{x ^{2} + 4} d x = \frac{7 π}{4} - \frac{7}{2} arctan (\frac{1}{2})

= \frac{7 π}{4} - \frac{7}{2} arctan (\frac{1}{2})

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