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受欢迎的微积分 >

integral of 9/(x(x^4+1))

解答

\int \frac{9}{x ( x ^{4} + 1 )} d x

解答

\frac{9}{4} (4 ln (x) - ln 1 + x^{4}) + C

求解步骤

\int \frac{9}{x ( x ^{4} + 1 )} d x

提出常数:

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

= 9 \cdot \int \frac{1}{x ( x ^{4} + 1 )} d x

使用换元积分法

= 9 \cdot \int \frac{1}{4 ( u - 1 ) u} d u

提出常数:

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

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{( u - 1 ) u} d u

将

\frac{1}{( u - 1 ) u}

用部份分式展开:

\frac{1}{u - 1} - \frac{1}{u}

= 9 \cdot \frac{1}{4} \cdot \int \frac{1}{u - 1} - \frac{1}{u} 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

= 9 \cdot \frac{1}{4} (\int \frac{1}{u - 1} d u - \int \frac{1}{u} d u)

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

= 9 \cdot \frac{1}{4} (ln ∣ u - 1 ∣ - ln ∣ u ∣)

u = 1 + x^{4}

代回

= 9 \cdot \frac{1}{4} (ln 1 + x^{4} - 1 - ln 1 + x^{4})

化简

9 \cdot \frac{1}{4} (ln 1 + x^{4} - 1 - ln 1 + x^{4}) : \frac{9}{4} (4 ln (x) - ln 1 + x^{4})

= \frac{9}{4} (4 ln (x) - ln 1 + x^{4})

解答补常数

= \frac{9}{4} (4 ln (x) - ln 1 + x^{4}) + C

作图

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