# $\rm \int {2x^3+4x\over x^4+5+4x^2}dx$ =

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