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Solve \(\rm \int {6x\over(3x^2+1)^2}\) dx1. \(\rm -{1\over3x^2+1}+c\)2. \(\rm {1\over3x^2+1}+c\)3. \(\rm -{1\over(3x^2+1)^3}+c\)4. \(\rm {3\over3x^2+1}+c\) |
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Answer» Correct Answer - Option 1 : \(\rm -{1\over3x^2+1}+c\) Concept: Integral property:
Substitution method: If the function cannot be integrated directly substitution method is used. To integration by substitution is used in the following steps:
Calculation: I = \(\rm \int {6x\over(3x^2+1)^2}\) Let x2 = t Differentiating both sides ⇒ 2x dx = dt Now, I = \(\rm \int {6x\over(3x^2+1)^2}\) dx ⇒ I = \(\rm 3\int {2x\over(3x^2+1)^2}\)dx Substituting x2 by t and dt = 2x dx ⇒ I = 3 \(\rm \int {1\over(3t+1)^2}\) dt ⇒ I = \(\rm 3\left[ {(3t+1)^{-1}\over-3}\right] + c\) ⇒ I = \(\rm {-1\over(3t+1)}\) + c Substituting t as x2 ⇒ I = \(\boldsymbol{\rm -{1\over3x^2+1}+c}\) |
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