integral from 0 to 1 of (x^2+x+1)/(x+1)

解答

\int_{0}^{1} \frac{x ^{2} + x + 1}{x + 1} d x

ln (2) + \frac{1}{2}

十进制

1.19314 \dots

求解步骤

\int_{0}^{1} \frac{x ^{2} + x + 1}{x + 1} d x

乘开

\frac{x ^{2} + x + 1}{x + 1} : \frac{x ^{2}}{x + 1} + \frac{x}{x + 1} + \frac{1}{x + 1}

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_{0}^{1} \frac{x ^{2}}{x + 1} d x + \int_{0}^{1} \frac{x}{x + 1} d x + \int_{0}^{1} \frac{1}{x + 1} d x

\int_{0}^{1} \frac{x ^{2}}{x + 1} d x = ln (2) - \frac{1}{2}

\int_{0}^{1} \frac{x}{x + 1} d x = 1 - ln (2)

\int_{0}^{1} \frac{1}{x + 1} d x = ln (2)

= ln (2) - \frac{1}{2} + 1 - ln (2) + ln (2)

化简

ln (2) - \frac{1}{2} + 1 - ln (2) + ln (2) : ln (2) + \frac{1}{2}

= ln (2) + \frac{1}{2}

\int_{- 2}^{x} f (t) d t

\int_{4}^{6} x^{2} + x - 9 d x

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

\int_{0}^{b} \frac{x}{x ^{2} + 1} d x

\int_{0}^{3} \frac{2 x + 3}{1 + x ^{2}} d x