integral from 0 to 1 of ln(4+x^2)

解答

\int_{0}^{1} ln (4 + x^{2}) d x

ln (5) - 4 (\frac{1}{2} - arctan (\frac{1}{2}))

十进制

1.46402 \dots

求解步骤

\int_{0}^{1} ln (4 + x^{2}) d x

使用分布积分法

= [x ln (4 + x^{2}) - \int \frac{2 x ^{2}}{4 + x ^{2}} d x]_{0}^{1}

\int \frac{2 x ^{2}}{4 + x ^{2}} d x = 4 (- arctan (\frac{x}{2}) + \frac{x}{2})

= [x ln (4 + x^{2}) - 4 (- arctan (\frac{x}{2}) + \frac{x}{2})]_{0}^{1}

带入上下限计算后得:

ln (5) - 4 (\frac{1}{2} - arctan (\frac{1}{2}))

= ln (5) - 4 (\frac{1}{2} - arctan (\frac{1}{2}))

\int - u cos (2 u) d u

\int \frac{1}{[ ( x ^{2} + 9 ) ^{2} ]} d x

\frac{d}{d x} (\frac{7 ^{x}}{x ^{3}})

\int \frac{e ^{2 x} - 1}{e ^{2 x} + 3} d x

x \to 3 lim (\frac{x + 1}{x ^{2} - 9})