integral from 1 to 4 of 7(ln(x))/(x^2)

解答

\int_{1}^{4} 7 \frac{ln ( x )}{x ^{2}} d x

7 (\frac{- 2 ln ( 2 ) - 1}{4} + 1)

十进制

2.82398 \dots

求解步骤

\int_{1}^{4} 7 \cdot \frac{ln ( x )}{x ^{2}} d x

提出常数:

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

= 7 \cdot \int_{1}^{4} \frac{ln ( x )}{x ^{2}} d x

使用分布积分法

= 7 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{4}

\int - \frac{1}{x ^{2}} d x = \frac{1}{x}

= 7 [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= 7 [\frac{- ln ( x ) - 1}{x}]_{1}^{4}

带入上下限计算后得:

\frac{- 2 ln ( 2 ) - 1}{4} + 1

= 7 (\frac{- 2 ln ( 2 ) - 1}{4} + 1)

\int \frac{1}{x + 8} d x

\int_{0}^{1} \frac{2 x}{( x ^{2} + 5 ) ^{2}} d x

\int_{0}^{\frac{π}{4}} (sin (x))^{4} d x

\int (3 u^{- 2} - 4 u^{2} + 1) d u

\int (3 x - 2)^{20} d x