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Miscellaneous Exercise 4 · Q51

Q.Choose the correct option from the given alternatives: ∫23dxx(x3−1)=\int_2^3 \dfrac{dx}{x(x^3-1)} = (A) 13log⁡208189\frac13\log\frac{208}{189} (B) 13log⁡189208\frac13\log\frac{189}{208} (C) log⁡208189\log\frac{208}{189} (D) log⁡189208\log\frac{189}{208}

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✓ Free question

1x(x3−1)=1x(x−1)(x2+x+1)=−1x+1/3x−1+(2/3)x+1/3x2+x+1\dfrac1{x(x^3-1)}=\dfrac1{x(x-1)(x^2+x+1)}=-\dfrac1x+\dfrac{1/3}{x-1}+\dfrac{(2/3)x+1/3}{x^2+x+1} (solving the partial-fraction identity by substituting x=0x=0 and x=1x=1, and matching the remaining coefficients).

The last piece integrates using 2x+1=ddx(x2+x+1)2x+1=\frac{d}{dx}(x^2+x+1): overall antiderivative =−ln⁡x+13ln⁡∣x−1∣+13ln⁡(x2+x+1)=-\ln x+\tfrac13\ln|x-1|+\tfrac13\ln(x^2+x+1).

Evaluating 22 to 33: at x=3x=3: −ln⁡3+13ln⁡2+13ln⁡13-\ln3+\tfrac13\ln2+\tfrac13\ln13. At x=2x=2: −ln⁡2+13ln⁡7-\ln2+\tfrac13\ln7 (since ln⁡1=0\ln1=0).

Difference =−ln⁡3+43ln⁡2+13ln⁡13−13ln⁡7=13[4ln⁡2−3ln⁡3+ln⁡13−ln⁡7]=13ln⁡16⋅1327⋅7=13ln⁡208189=-\ln3+\tfrac43\ln2+\tfrac13\ln13-\tfrac13\ln7=\tfrac13\big[4\ln2-3\ln3+\ln13-\ln7\big]=\tfrac13\ln\dfrac{16\cdot13}{27\cdot7}=\tfrac13\ln\dfrac{208}{189}.

✓Final answer

∫23dxx(x3−1)=13log⁡208189\displaystyle\int_2^3\frac{dx}{x(x^3-1)}=\frac13\log\frac{208}{189} — option (A)

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