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受欢迎的微积分 >

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

解答

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

解答

4 (\frac{1}{32} ln 2 x + 4 + 2 - \frac{1}{16 ( 2 x + 4 + 2 )} - \frac{1}{32} ln 2 x + 4 - 2 - \frac{1}{16 ( 2 x + 4 - 2 )}) + C

求解步骤

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

使用指数法则:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{x ^{2} 1 + 2 x + 3 x ^{2}} = x^{- 2} \frac{1}{1 + 2 x + 3}

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

化简

x^{- 2} \frac{1}{1 + 2 x + 3} : \frac{1}{x ^{2} 2 x + 4}

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

使用换元积分法

= \int \frac{4}{( u ^{2} - 4 ) ^{2}} 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}{( u ^{2} - 4 ) ^{2}} d u

将

\frac{1}{( u ^{2} - 4 ) ^{2}}

用部份分式展开:

\frac{1}{32 ( u + 2 )} + \frac{1}{16 ( u + 2 ) ^{2}} - \frac{1}{32 ( u - 2 )} + \frac{1}{16 ( u - 2 ) ^{2}}

= 4 \cdot \int \frac{1}{32 ( u + 2 )} + \frac{1}{16 ( u + 2 ) ^{2}} - \frac{1}{32 ( u - 2 )} + \frac{1}{16 ( u - 2 ) ^{2}} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{1}{32 ( u + 2 )} d u + \int \frac{1}{16 ( u + 2 ) ^{2}} d u - \int \frac{1}{32 ( u - 2 )} d u + \int \frac{1}{16 ( u - 2 ) ^{2}} d u)

\int \frac{1}{32 ( u + 2 )} d u = \frac{1}{32} ln ∣ u + 2 ∣

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

\int \frac{1}{32 ( u - 2 )} d u = \frac{1}{32} ln ∣ u - 2 ∣

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

= 4 (\frac{1}{32} ln ∣ u + 2 ∣ - \frac{1}{16 ( u + 2 )} - \frac{1}{32} ln ∣ u - 2 ∣ - \frac{1}{16 ( u - 2 )})

u = 2 x + 4

代回

= 4 (\frac{1}{32} ln 2 x + 4 + 2 - \frac{1}{16 ( 2 x + 4 + 2 )} - \frac{1}{32} ln 2 x + 4 - 2 - \frac{1}{16 ( 2 x + 4 - 2 )})

解答补常数

= 4 (\frac{1}{32} ln 2 x + 4 + 2 - \frac{1}{16 ( 2 x + 4 + 2 )} - \frac{1}{32} ln 2 x + 4 - 2 - \frac{1}{16 ( 2 x + 4 - 2 )}) + C

作图

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