integral from 1 to 7 of 1/(36+(x+5)^2)

解答

\int_{1}^{7} \frac{1}{36 + ( x + 5 ) ^{2}} d x

\frac{1}{6} (arctan (2) - \frac{π}{4})

十进制

0.05362 \dots

求解步骤

\int_{1}^{7} \frac{1}{36 + ( x + 5 ) ^{2}} d x

使用换元积分法

= \int_{6}^{12} \frac{1}{36 + u ^{2}} d u

使用换元积分法

= \int_{1}^{2} \frac{1}{6 ( v ^{2} + 1 )} d v

提出常数:

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

= \frac{1}{6} \cdot \int_{1}^{2} \frac{1}{v ^{2} + 1} d v

使用积分常见定则:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{6} [arctan (v)]_{1}^{2}

带入上下限计算后得:

arctan (2) - \frac{π}{4}

= \frac{1}{6} (arctan (2) - \frac{π}{4})

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

\int (- 2 x^{4} - 4 x^{2} - 5) d x

\int 2 cos (\frac{x}{2}) d x

\frac{\partial}{\partial y} (2 ln (x) + ln (y))

\frac{\partial}{\partial x} (\frac{x}{y})