integral from 0 to 4 of x/(1+sqrt(2x+1))

解答

\int_{0}^{4} \frac{x}{1 + 2 x + 1} d x

\frac{7}{3}

十进制

2.33333 \dots

求解步骤

\int_{0}^{4} \frac{x}{1 + 2 x + 1} d x

使用换元积分法

= \int_{1}^{9} \frac{u - 1}{4 ( 1 + u )} d u

提出常数:

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

= \frac{1}{4} \cdot \int_{1}^{9} \frac{u - 1}{1 + u} d u

\frac{u - 1}{1 + u} = u - 1

= \frac{1}{4} \cdot \int_{1}^{9} u - 1 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}{4} (\int_{1}^{9} u d u - \int_{1}^{9} 1 d u)

\int_{1}^{9} u d u = \frac{52}{3}

\int_{1}^{9} 1 d u = 8

= \frac{1}{4} (\frac{52}{3} - 8)

化简

\frac{1}{4} (\frac{52}{3} - 8) : \frac{7}{3}

= \frac{7}{3}

\int_{5}^{7} x^{2} - 2 x - 15 d x

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