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Evaluate \(\rm \int \frac{x}{3x^2+4} dx \)1. \(\rm \frac{1}{3}\log (3x^2+4)+ c\)2. \(\rm \frac{1}{6}\log (3x^2+4)+ c\)3. \(\rm \frac{1}{6}\log (3x^2+4)+\tan^{-1} x + c\)4. None of the above |
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Answer» Correct Answer - Option 2 : \(\rm \frac{1}{6}\log (3x^2+4)+ c\) Concept: \(\rm \int \frac{1}{x} dx = \log x + c\) Calculation: I = \(\rm \int \frac{x}{3x^2+4} dx \) Let 3x2 + 4 = t Differentiating with respect to x, we get ⇒ 6xdx = dt ⇒ xdx = \(\rm \frac {dt}{6}\) Now, I = \(\rm \frac{1}{6}\int \frac{1}{t} dt\) = \(\rm \frac{1}{6}\log t + c\) = \(\rm \frac{1}{6}\log (3x^2+4)+ c\) |
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