将 2x3+x2+x(x−1)(x2+1)\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} 化为部分分式。 求 ln⁡(x2+1)\ln(x^2 + 1) 对 xx 的导数。 由此,求 ∫2x3+x2+x(x−1)(x2+1) dx\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx。

先不看答案,让 AI 诊断你卡在哪里?

它不会直接告诉你答案,而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

7(a)

1. Recognise that the fraction is improper and perform long division to extract the quotient.

The degree of the numerator is 3 and the degree of the denominator is 3, so the fraction is improper.
Denominator: (x−1)(x2+1)=x3−x2+x−1(x-1)(x^2 + 1) = x^3 - x^2 + x - 1

By long division or inspection:
2x3+x2+xx3−x2+x−1=2+(2x3+x2+x)−2(x3−x2+x−1)(x−1)(x2+1)\frac{2x^3 + x^2 + x}{x^3 - x^2 + x - 1} = 2 + \frac{(2x^3 + x^2 + x) - 2(x^3 - x^2 + x - 1)}{(x-1)(x^2 + 1)}
=2+3x2−x+2(x−1)(x2+1)= 2 + \frac{3x^2 - x + 2}{(x-1)(x^2 + 1)}

2. Set up the partial fraction decomposition for the proper fraction part.

Let 3x2−x+2(x−1)(x2+1)=Ax−1+Bx+Cx2+1\frac{3x^2 - x + 2}{(x-1)(x^2 + 1)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 1}

Multiplying by (x−1)(x2+1)(x-1)(x^2 + 1) gives:
3x2−x+2=A(x2+1)+(Bx+C)(x−1)3x^2 - x + 2 = A(x^2 + 1) + (Bx + C)(x - 1)

3. Solve for the constants $A$, $B$, and $C$.

Substitute x=1x = 1:
3(1)2−1+2=A(12+1)4=2A  ⟹  A=23(1)^2 - 1 + 2 = A(1^2 + 1) 4 = 2A \implies A = 2

Substitute x=0x = 0:
2=A(1)+C(−1)2=2−C  ⟹  C=02 = A(1) + C(-1) 2 = 2 - C \implies C = 0

Compare coefficients of x2x^2:
3=A+B3=2+B  ⟹  B=13 = A + B 3 = 2 + B \implies B = 1

4. Write the final expression in partial fractions.

2x3+x2+x(x−1)(x2+1)=2+2x−1+xx2+1\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} = 2 + \frac{2}{x-1} + \frac{x}{x^2 + 1}

答案

2+2x−1+xx2+12 + \frac{2}{x-1} + \frac{x}{x^2 + 1}

7(b)

1. Differentiate the function using the chain rule

ddx[ln⁡(x2+1)]=1x2+1⋅ddx(x2+1)\frac{d}{dx}[\ln(x^2 + 1)] = \frac{1}{x^2 + 1} \cdot \frac{d}{dx}(x^2 + 1)

2. Simplify the derivative

=1x2+1⋅(2x)=2xx2+1= \frac{1}{x^2 + 1} \cdot (2x) = \frac{2x}{x^2 + 1}

答案

2xx2+1\frac{2x}{x^2 + 1}

7(c)

1. Express the integrand using the partial fractions found in part (a)

From part (a), we have:
2x3+x2+x(x−1)(x2+1)=2+2x−1+xx2+1\frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} = 2 + \frac{2}{x-1} + \frac{x}{x^2+1}

2. Set up the integral using the partial fractions

∫2x3+x2+x(x−1)(x2+1) dx=∫(2+2x−1+xx2+1) dx\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx = \int \left( 2 + \frac{2}{x-1} + \frac{x}{x^2+1} \right) \, dx

3. Integrate the first two terms

∫2 dx=2x\int 2 \, dx = 2x
∫2x−1 dx=2ln⁡∣x−1∣\int \frac{2}{x-1} \, dx = 2\ln|x-1|

4. Use the result from part (b) to integrate the last term and state the final answer

From part (b), ddxln⁡(x2+1)=2xx2+1\frac{d}{dx} \ln(x^2+1) = \frac{2x}{x^2+1}.
Therefore, ∫xx2+1 dx=12ln⁡(x2+1)\int \frac{x}{x^2+1} \, dx = \frac{1}{2}\ln(x^2+1).

Combining all parts:
∫2x3+x2+x(x−1)(x2+1) dx=2x+2ln⁡∣x−1∣+12ln⁡(x2+1)+c\int \frac{2x^3 + x^2 + x}{(x-1)(x^2 + 1)} \, dx = 2x + 2\ln|x-1| + \frac{1}{2}\ln(x^2+1) + c

答案

2x+2ln⁡∣x−1∣+12ln⁡(x2+1)+c2x + 2\ln|x-1| + \frac{1}{2}\ln(x^2+1) + c

AI 分析,可能出错

用 AI 批改,圈出写错的那一步,并自动将你的薄弱点存入错题本,形成你的错因库。

同一知识点的其他题

  1. 1将 2x4+x3+9x2+7x+3(x+1)(x2+5)\frac{2x^4 + x^3 + 9x^2 + 7x + 3}{(x+1)(x^2+5)} 化为部分分式。
  2. 2将 x2+3x+2x2(2x+1)\dfrac{x^2+3x+2}{x^2(2x+1)} 表示为部分分式。 由此对 x2+3x+2x2(2x+1)\dfrac{x^2+3x+2}{x^2(2x+1)} 关于 xx 进行积分。
  3. 3将 8x+13(1+2x)(2+x)2\frac{8x+13}{(1+2x)(2+x)^2} 表示为部分分式。 将 8x+13(1+2x)(2+x)2\frac{8x+13}{(1+2x)(2+x)^2} 表示为部分分式。 由此,求 ∫128x+13(1+2x)(2+x)2 dx\displaystyle \int_{1}^{2} \frac{8x+13}{(1+2x)(2+x)^2}\,\mathrm{d}x…
  4. 4将 2x2+3xx2+3x+2\dfrac{2x^{2}+3x}{x^{2}+3x+2} 表示为部分分式。 由此计算 ∫132x2+3xx2+3x+2 dx\displaystyle \int_{1}^{3} \dfrac{2x^{2}+3x}{x^{2}+3x+2}\,dx。 已知 y=xln⁡(x2+3x+2)y = x\ln\left(x^{2}+3x+2\right)…
  5. 5将 3x2+5x−2(x+1)(x2+4)\frac{3x^2 + 5x - 2}{(x + 1)(x^2 + 4)} 化为部分分式。