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受欢迎的微积分 >

integral of 1/(6x^2+3x+1)

解答

\int \frac{1}{6 x ^{2} + 3 x + 1} d x

解答

\frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5}) + C

求解步骤

\int \frac{1}{6 x ^{2} + 3 x + 1} d x

对

6 x^{2} + 3 x + 1

配方:

6 (x + \frac{1}{4})^{2} + \frac{5}{8}

= \int \frac{1}{6 ( x + \frac{1}{4} ) ^{2} + \frac{5}{8}} d x

使用换元积分法

= \int \frac{8}{48 u ^{2} + 5} 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}{48 u ^{2} + 5} d u

使用换元积分法

= 8 \cdot \int \frac{1}{4 15 ( v ^{2} + 1 )} d v

提出常数:

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

= 8 \cdot \frac{1}{4 15} \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)

= 8 \cdot \frac{1}{4 15} arctan (v)

代回

= 8 \cdot \frac{1}{4 15} arctan (\frac{4 3}{5} (x + \frac{1}{4}))

化简

8 \cdot \frac{1}{4 15} arctan (\frac{4 3}{5} (x + \frac{1}{4})) : \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5})

= \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5})

解答补常数

= \frac{2}{15} arctan (\frac{4 3}{5} x + \frac{3}{5}) + C

作图

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