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∫9+(log⁡x)2xdx\int \frac{\sqrt{9+(\log x)^2}}{x} dx is equal to: (where CC is an arbitrary constant and consider log⁡ex=log⁡x\log_e x = \log x)

Solution

✅ Correct Option: 2

Put t=log⁡xt=\log x, so dt=dxxdt=\frac{dx}{x} and the integral becomes ∫9+t2 dt\int\sqrt{9+t^2}\,dt.

Using ∫t2+a2 dt=t2t2+a2+a22log⁡∣t+t2+a2∣+C\int\sqrt{t^2+a^2}\,dt=\frac{t}{2}\sqrt{t^2+a^2}+\frac{a^2}{2}\log|t+\sqrt{t^2+a^2}|+C with a=3a=3, we get 12log⁡x9+(log⁡x)2+92log⁡∣log⁡x+9+(log⁡x)2∣+C\frac{1}{2}\log x\sqrt{9+(\log x)^2}+\frac{9}{2}\log|\log x+\sqrt{9+(\log x)^2}|+C.

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