integral from 1 to 3 of 3/(5x^2+5)

解答

\int_{1}^{3} \frac{3}{5 x ^{2} + 5} d x

\frac{3}{5} (arctan (3) - \frac{π}{4})

十进制

0.27818 \dots

求解步骤

\int_{1}^{3} \frac{3}{5 x ^{2} + 5} d x

提出常数:

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

= 3 \cdot \int_{1}^{3} \frac{1}{5 x ^{2} + 5} d x

\frac{1}{5 x ^{2} + 5} = \frac{1}{5} \frac{1}{x ^{2} + 1}

= 3 \cdot \int_{1}^{3} \frac{1}{5} \cdot \frac{1}{x ^{2} + 1} d x

提出常数:

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

= 3 \cdot \frac{1}{5} \cdot \int_{1}^{3} \frac{1}{x ^{2} + 1} d x

使用积分常见定则:

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

= 3 \cdot \frac{1}{5} [arctan (x)]_{1}^{3}

化简

3 \cdot \frac{1}{5} [arctan (x)]_{1}^{3} : \frac{3}{5} [arctan (x)]_{1}^{3}

= \frac{3}{5} [arctan (x)]_{1}^{3}

带入上下限计算后得:

arctan (3) - \frac{π}{4}

= \frac{3}{5} (arctan (3) - \frac{π}{4})

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