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

解答

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

\frac{2 arctan ( \frac{2}{2} ) + π}{2 2}

十进制

1.54593 \dots

求解步骤

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

计算不定积分:

\int \frac{1}{x ^{2} + 2} d x = \frac{1}{2} arctan (\frac{x}{2}) + C

带入上下限计算后得:

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

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

化简

= \frac{2 arctan ( \frac{2}{2} ) + π}{2 2}

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