Maths · Integration by substitution, by parts and by partial fractions
\( \frac{\boldsymbol{x}+\mathbf{2}}{\boldsymbol{x}^{\boldsymbol{3}}-\boldsymbol{x}}= \)
- A. \( \frac{1}{2(x+1)}+\frac{3}{2(x-1)}-\frac{2}{x} \)
- B. \( \frac{1}{2(x+1)}-\frac{3}{2(x-1)}-\frac{2}{x} \)
- C. \( \frac{1}{2(x+1)}-\frac{3}{2(x-1)}+\frac{2}{x} \)
- D. \( \frac{1}{2(x+1)}+\frac{3}{2(x-1)}+\frac{2}{x} \)
Step-by-step solution
Factor denominator: \(x^3 - x = x(x-1)(x+1)\). Write partial fractions:
\[
\frac{x+2}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}.
\]
Multiply both sides by denominator: \(x+2 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)\).
Substitute \(x=0\): \(2 = -A \Rightarrow A = -2\).
Substitute \(x=1\): \(3 = 2B \Rightarrow B = \frac{3}{2}\).
Substitute \(x=-1\): \(1 = 2C \Rightarrow C = \frac{1}{2}\).
Thus decomposition: \(-\frac{2}{x} + \frac{3}{2(x-1)} + \frac{1}{2(x+1)}\). Rearranged matches option A.
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