integral from y to 8-y of 3/(x^2+y^2)

解答

\int_{y}^{8 - y} \frac{3}{x ^{2} + y ^{2}} d y

\frac{3 ( arctan ( \frac{8 - y}{x} ) - arctan ( \frac{y}{x} ) )}{x}

求解步骤

\int_{y}^{8 - y} \frac{3}{x ^{2} + y ^{2}} d y

提出常数:

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

= 3 \cdot \int_{y}^{8 - y} \frac{1}{x ^{2} + y ^{2}} d y

使用换元积分法

= 3 \cdot \int_{\frac{y}{x}}^{\frac{8 - y}{x}} \frac{1}{x ( u ^{2} + 1 )} d u

提出常数:

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

= 3 \cdot \frac{1}{x} \cdot \int_{\frac{y}{x}}^{\frac{8 - y}{x}} \frac{1}{u ^{2} + 1} d u

使用积分常见定则:

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

= 3 \cdot \frac{1}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

化简

3 \cdot \frac{1}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}} : \frac{3}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

= \frac{3}{x} [arctan (u)]_{\frac{y}{x}}^{\frac{8 - y}{x}}

带入上下限计算后得:

arctan (\frac{8 - y}{x}) - arctan (\frac{y}{x})

= \frac{3}{x} (arctan (\frac{8 - y}{x}) - arctan (\frac{y}{x}))

化简

= \frac{3 ( arctan ( \frac{8 - y}{x} ) - arctan ( \frac{y}{x} ) )}{x}

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