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受欢迎的微积分 >

integral of 1/(t^2-t+2)

解答

\int \frac{1}{t ^{2} - t + 2} d t

解答

\frac{2}{7} arctan (\frac{2 t - 1}{7}) + C

求解步骤

\int \frac{1}{t ^{2} - t + 2} d t

对

t^{2} - t + 2

配方:

(t - \frac{1}{2})^{2} + \frac{7}{4}

= \int \frac{1}{( t - \frac{1}{2} ) ^{2} + \frac{7}{4}} d t

使用换元积分法

= \int \frac{4}{4 u ^{2} + 7} d u

提出常数:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 7} d u

使用换元积分法

= 4 \cdot \int \frac{1}{2 7 ( v ^{2} + 1 )} d v

提出常数:

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

= 4 \cdot \frac{1}{2 7} \cdot \int \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)

= 4 \cdot \frac{1}{2 7} arctan (v)

代回

= 4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (t - \frac{1}{2}))

化简

4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (t - \frac{1}{2})) : \frac{2}{7} arctan (\frac{2 t - 1}{7})

= \frac{2}{7} arctan (\frac{2 t - 1}{7})

解答补常数

= \frac{2}{7} arctan (\frac{2 t - 1}{7}) + C

作图

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