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受欢迎的微积分 >

integral of 1/(2x^2-6x+5)

解答

\int \frac{1}{2 x ^{2} - 6 x + 5} d x

解答

arctan (2 x - 3) + C

求解步骤

\int \frac{1}{2 x ^{2} - 6 x + 5} d x

对

2 x^{2} - 6 x + 5

配方:

2 (x - \frac{3}{2})^{2} + \frac{1}{2}

= \int \frac{1}{2 ( x - \frac{3}{2} ) ^{2} + \frac{1}{2}} d x

使用换元积分法

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

提出常数:

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

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

使用换元积分法

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

提出常数:

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

= 2 \cdot \frac{1}{2} \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)

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

代回

= 2 \cdot \frac{1}{2} arctan (2 (x - \frac{3}{2}))

化简

2 \cdot \frac{1}{2} arctan (2 (x - \frac{3}{2})) : arctan (2 x - 3)

= arctan (2 x - 3)

解答补常数

= arctan (2 x - 3) + C

作图

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