integral from 4 to 5 of (4x^2)/(x-3)

解答

\int_{4}^{5} \frac{4 x ^{2}}{x - 3} d x

4 (9 ln (2) + \frac{15}{2})

十进制

54.95329 \dots

求解步骤

\int_{4}^{5} \frac{4 x ^{2}}{x - 3} d x

提出常数:

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

= 4 \cdot \int_{4}^{5} \frac{x ^{2}}{x - 3} d x

使用换元积分法

= 4 \cdot \int_{1}^{2} \frac{( u + 3 ) ^{2}}{u} d u

乘开

\frac{( u + 3 ) ^{2}}{u} : u + 6 + \frac{9}{u}

= 4 \cdot \int_{1}^{2} u + 6 + \frac{9}{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

= 4 (\int_{1}^{2} u d u + \int_{1}^{2} 6 d u + \int_{1}^{2} \frac{9}{u} d u)

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

\int_{1}^{2} 6 d u = 6

\int_{1}^{2} \frac{9}{u} d u = 9 ln (2)

= 4 (\frac{3}{2} + 6 + 9 ln (2))

合并分式

6 + \frac{3}{2} : \frac{15}{2}

= 4 (9 ln (2) + \frac{15}{2})

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