integral from 0 to y of x/(x^2+y^2)

解答

\int_{0}^{y} \frac{x}{x ^{2} + y ^{2}} d y

arctan (\frac{y}{x})

求解步骤

\int_{0}^{y} \frac{x}{x ^{2} + y ^{2}} d y

提出常数:

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

= x \cdot \int_{0}^{y} \frac{1}{x ^{2} + y ^{2}} d y

使用换元积分法

= x \cdot \int_{0}^{\frac{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

= x \frac{1}{x} \cdot \int_{0}^{\frac{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)

= x \frac{1}{x} [arctan (u)]_{0}^{\frac{y}{x}}

化简

x \frac{1}{x} [arctan (u)]_{0}^{\frac{y}{x}} : [arctan (u)]_{0}^{\frac{y}{x}}

= [arctan (u)]_{0}^{\frac{y}{x}}

带入上下限计算后得:

arctan (\frac{y}{x})

= arctan (\frac{y}{x})

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