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受欢迎的微积分 >

integral of ((12x^2))/(2x+1)

解答

\int \frac{( 12 x ^{2} )}{2 x + 1} d x

解答

\frac{3}{4} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) + C

求解步骤

\int \frac{12 x ^{2}}{2 x + 1} d x

提出常数:

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

= 12 \cdot \int \frac{x ^{2}}{2 x + 1} d x

使用换元积分法

= 12 \cdot \int \frac{( u - 1 ) ^{2}}{8 u} d u

提出常数:

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

= 12 \cdot \frac{1}{8} \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

乘开

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

= 12 \cdot \frac{1}{8} \cdot \int u - 2 + \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

= 12 \cdot \frac{1}{8} (\int u d u - \int 2 d u + \int \frac{1}{u} d u)

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= 12 \cdot \frac{1}{8} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

u = 2 x + 1

代回

= 12 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣)

化简

12 \cdot \frac{1}{8} (\frac{( 2 x + 1 ) ^{2}}{2} - 2 (2 x + 1) + ln ∣ 2 x + 1 ∣) : \frac{3}{4} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

= \frac{3}{4} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3)

解答补常数

= \frac{3}{4} (4 x^{2} - 4 x + 2 ln ∣ 2 x + 1 ∣ - 3) + C

作图

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