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\(\rm \int {2x^3+4x\over x^4+5+4x^2}dx\) = 
1. \(\rm \ln (x^2 + 4)+c\)
2. \(\rm {1\over2}\ln (x^2+2)+c\)
3. \(\rm \ln (x^4 + 4x^2 + 5)+c\)
4. \(\rm {1\over2}\ln (x^4 + 4x^2 + 5)+c\)
5. \(\rm \ln (x^2 + 2)+c\)

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Correct Answer - Option 4 : \(\rm {1\over2}\ln (x^4 + 4x^2 + 5)+c\)

Concept:

Integral property:

  • ∫ xn dx = \(\rm x^{n+1}\over n+1\)+ C ; n ≠ -1
  • \(\rm∫ {1\over x} dx = \ln x\) + C
  • ∫ edx = ex+ C
  • ∫ adx = (ax/ln a) + C ; a > 0,  a ≠ 1
  • ∫ sin x dx = - cos x + C
  • ∫ cos x dx = sin x + C


Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:

  • A new variable is to be chosen, say “t”
  • The value of dt is to be determined.
  • Substitution is done and the integral function is then integrated.
  • Finally, the initial variable t, to be returned.
 

Calculation:

I = \(\rm \int {2x^3+4x\over x^4+5+4x^2}dx\)

I = \(\rm \int {2x^3+4x\over x^4+4x^2 +5}dx\)

I = \(\rm \frac 1 2 \int {4x^3+8x\over x^4+4x^2 +5}dx\)

Let x4 + 4x2 = t

⇒ (4x3 + 8x)dx = dt

Now,

I = \(\rm {1\over2}\int {1\over t}dt\)

I = \(\rm {1\over2}\ln t + c\)

I = \(\boldsymbol{\rm {1\over2}\ln (x^4 + 4x^2 + 5)+c}\)

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