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

解答

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

\frac{2 ln ( 2 ) - 1}{16}

十进制

0.02414 \dots

求解步骤

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

使用换元积分法

= \int_{1}^{2} \frac{( u - 1 ) ^{2}}{8 u} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int_{1}^{2} \frac{( u - 1 ) ^{2}}{u} d u

乘开

\frac{( u - 1 ) ^{2}}{u} : u - 2 + \frac{1}{u}

= \frac{1}{8} \cdot \int_{1}^{2} u - 2 + \frac{1}{u} d u

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

= \frac{1}{8} (\int_{1}^{2} u d u - \int_{1}^{2} 2 d u + \int_{1}^{2} \frac{1}{u} d u)

\int_{1}^{2} u d u = \frac{3}{2}

\int_{1}^{2} 2 d u = 2

\int_{1}^{2} \frac{1}{u} d u = ln (2)

= \frac{1}{8} (\frac{3}{2} - 2 + ln (2))

化简

\frac{1}{8} (\frac{3}{2} - 2 + ln (2)) : \frac{2 ln ( 2 ) - 1}{16}

= \frac{2 ln ( 2 ) - 1}{16}

\int_{3}^{2} (x) d x

\int_{0}^{2} (5 x cos (π x)) d x

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

\int_{0}^{4} \frac{x}{x ^{2} - 3 x} d x

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