Find the integral of 3e^x+\frac{2(log⁡ x)}{3x}.

Category: QuestionsFind the integral of 3e^x+\frac{2(log⁡ x)}{3x}.
Editor">Editor Staff asked 11 months ago

Find the integral of 3e^x+\frac{2(log⁡ x)}{3x}.
 
(a) 3e^x+\frac{1}{3} (x)^2+C
 
(b) e^x-\frac{8}{3} (log⁡x)^2+C
 
(c) 3e^x-\frac{1}{3} (log⁡x)^2+C
 
(d) 3e^x+\frac{1}{3} (log⁡x)^2+C
 
I have been asked this question in a national level competition.
 
Query is from Methods of Integration-1 in section Integrals of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

The correct answer is (d) 3e^x+\frac{1}{3} (log⁡x)^2+C
 
Explanation: \int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3\int e^x dx+\frac{2}{3} \int \frac{log⁡x}{x}
 
Let log⁡x=t
 
Differentiating w.r.t x, we get
 
\frac{1}{x} dx=dt
 
∴\int \frac{log⁡x}{x}=\int \,t \,dt=\frac{t^2}{2}
 
\int e^x dx=e^x
 
Replacing t with log⁡x, we get
 
\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3e^x+\frac{1}{3} (log⁡x)^2+C