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受欢迎的微积分 >

integral of 1/(2x^2+3x+1)

解答

\int \frac{1}{2 x ^{2} + 3 x + 1} d x

解答

- ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣ + C

求解步骤

\int \frac{1}{2 x ^{2} + 3 x + 1} d x

对

2 x^{2} + 3 x + 1

配方:

2 (x + \frac{3}{4})^{2} - \frac{1}{8}

= \int \frac{1}{2 ( x + \frac{3}{4} ) ^{2} - \frac{1}{8}} d x

使用换元积分法

= \int \frac{8}{16 u ^{2} - 1} d u

提出常数:

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

= 8 \cdot \int \frac{1}{16 u ^{2} - 1} d u

因式分解

16 u^{2} - 1 : - (- 16 u^{2} + 1)

= 8 \cdot \int \frac{1}{- ( - 16 u ^{2} + 1 )} d u

提出常数:

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

= 8 (- \int \frac{1}{- 16 u ^{2} + 1} d u)

使用换元积分法

= 8 (- \int \frac{1}{4 ( - v ^{2} + 1 )} d v)

提出常数:

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

= 8 (- \frac{1}{4} \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 = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 (- \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

= 8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{3}{4} ) - 1}{2}))

化简

8 (- \frac{1}{4} (\frac{ln 4 ( x + \frac{3}{4} ) + 1}{2} - \frac{ln 4 ( x + \frac{3}{4} ) - 1}{2})) : - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣

= - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣

解答补常数

= - ln ∣ 4 x + 4 ∣ + ln ∣ 4 x + 2 ∣ + C

作图

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