integral of (1+x)

解答

\int (1 + x) d x

x + \frac{x ^{2}}{2} + C

求解步骤

\int 1 + x 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 d x + \int x d x

\int 1 d x = x

\int x d x = \frac{x ^{2}}{2}

= x + \frac{x ^{2}}{2}

解答补常数

= x + \frac{x ^{2}}{2} + C

\int_{3}^{2} \frac{x ^{2} - 1}{x - 1} d x

n = 1 \sum \infty (x - 6)^{n}

x \to + 1 - lim (\frac{1}{x ^{2} + 1})

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