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受欢迎的微积分 >

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

解答

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

解答

\frac{2}{19} arctan (\frac{2 x + 1}{19}) + C

求解步骤

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

对

5 + x + x^{2}

配方:

(x + \frac{1}{2})^{2} + \frac{19}{4}

= \int \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{19}{4}} d x

使用换元积分法

= \int \frac{4}{4 u ^{2} + 19} 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} + 19} d u

使用换元积分法

= 4 \cdot \int \frac{1}{2 19 ( 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 19} \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 19} arctan (v)

代回

= 4 \cdot \frac{1}{2 19} arctan (\frac{2}{19} (x + \frac{1}{2}))

化简

4 \cdot \frac{1}{2 19} arctan (\frac{2}{19} (x + \frac{1}{2})) : \frac{2}{19} arctan (\frac{2 x + 1}{19})

= \frac{2}{19} arctan (\frac{2 x + 1}{19})

解答补常数

= \frac{2}{19} arctan (\frac{2 x + 1}{19}) + C

作图

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