Question

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

Answer 1

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

\(\int 3e^x+\frac{2(log⁡x^2)}{3x} dx=3e^x+\frac{1}{3} (log⁡x)^2+C\)