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受欢迎的微积分 >

integral of 4/(4x^2-4x+5)

解答

\int \frac{4}{4 x ^{2} - 4 x + 5} d x

解答

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

求解步骤

\int \frac{4}{4 x ^{2} - 4 x + 5} d x

提出常数:

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

= 4 \cdot \int \frac{1}{4 x ^{2} - 4 x + 5} d x

对

4 x^{2} - 4 x + 5

配方:

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

= 4 \cdot \int \frac{1}{4 ( x - \frac{1}{2} ) ^{2} + 4} d x

使用换元积分法

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

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

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

提出常数:

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

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

= 4 \cdot \frac{1}{4} arctan (u)

u = x - \frac{1}{2}

代回

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

化简

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

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

解答补常数

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

作图

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