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\(\rm\int{dx\over x^5 +x^3}\) = ?
1. \(\rm -{1\over2x^2}-\log x+{1\over2}\log(x^2+1)\) + C
2. \(\rm {1\over2x^2}-\log x+{1\over2}\log(x^2+1)\) + C
3. \(\rm {1\over2x^2}+\log x-{1\over2}\log(x^2+1)\) + C
4. \(\rm {1\over2x^2}+2\log x-{1\over2}\log(x^2+1) + C\)
5. None of these

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Correct Answer - Option 1 : \(\rm -{1\over2x^2}-\log x+{1\over2}\log(x^2+1)\) + C

Concept:

Integral property:

  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C

Calculation:

I = \(\rm\int{dx\over x^5 +x^3}\)

I = \(\rm\int{dx\over x^3(x^2 +1)}\) 

By partial fraction

I = \(\rm \int {x\over x^2+1}-{1\over x}+{1\over x^3}\) dx

I = \(\rm \int {x\over x^2+1}dx-\int {dx\over x}+\int{dx\over x^3}\)

Let x2 + 1 = t ⇒ 2x dx = dt

I = \(\rm \int {dt\over2t}-\ln x+\int x^{-3}dx + C\)

I = \(\rm {1\over2}\ln t-\ln x+{x^{-2}\over-2} + C\)

Putting the value of t

I = \(\rm {1\over2}\ln (x^2+1)-\ln x-{1\over 2x^2} + C\)

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