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受欢迎的微积分 >

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

解答

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

解答

ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

求解步骤

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

提出常数:

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

= 2 \cdot \int \frac{x ^{3}}{( x ^{2} + 1 ) ^{2}} d x

使用换元积分法

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

提出常数:

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

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

乘开

\frac{u - 1}{u ^{2}} : \frac{1}{u} - \frac{1}{u ^{2}}

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 2 \cdot \frac{1}{2} (ln ∣ u ∣ - (- \frac{1}{u}))

u = x^{2} + 1

代回

= 2 \cdot \frac{1}{2} (ln x^{2} + 1 - (- \frac{1}{x ^{2} + 1}))

化简

2 \cdot \frac{1}{2} (ln x^{2} + 1 - (- \frac{1}{x ^{2} + 1})) : ln x^{2} + 1 + \frac{1}{x ^{2} + 1}

= ln x^{2} + 1 + \frac{1}{x ^{2} + 1}

解答补常数

= ln x^{2} + 1 + \frac{1}{x ^{2} + 1} + C

作图

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