Question

The value of integral \(\int \frac{\sqrt{x^2+1}\,[log(x^2+1)-2logx]}{x^4}\,dx\) is equal to...

• A. \((1+\frac{1}{x^2})^{3/2}[\frac{-1}{3}log(1+\frac{1}{x^2})+\frac{2}{9}]+c\)
• B. \((1+\frac{1}{x^2})^{3/2}[\frac{-1}{3}log(1+\frac{1}{x^2})-\frac{2}{9}]+c\)
• C. \((1+\frac{1}{x^2})^{3/2}[\frac{-1}{3}log(1+\frac{1}{x^2})+\frac{2}{3}]+c\)
• D. \((1+\frac{1}{x^2})^{3/2}[\frac{-1}{3}log(1+\frac{1}{x^2})-\frac{2}{3}]+c\)

Hint:

Rewrite the bracket as log(1 + 1/x^2) and substitute t = 1 + 1/x^2.

Answer 1

The Correct Option is A

Step 1: Simplify the Log Part:
\(\log(x^2+1)-2\log x=\log\dfrac{x^2+1}{x^2}=\log\left(1+\dfrac1{x^2}\right)\).

Step 2: Substitute:
Let \(t=1+\dfrac1{x^2}\). Then \(dt=-\dfrac{2}{x^3}dx\). Also \(\dfrac{\sqrt{x^2+1}}{x^4}dx=\dfrac{\sqrt{x^2+1}}{x}\cdot\dfrac{dx}{x^3}=\sqrt t\cdot\left(-\dfrac{dt}{2}\right)\).
\[ I=-\frac12\int\sqrt t\,\log t\,dt \]

Step 3: Integrate by Parts:
Take \(u=\log t\) and \(dv=t^{1/2}dt\), so \(v=\tfrac23t^{3/2}\).
\[ \int t^{1/2}\log t\,dt=\frac23t^{3/2}\log t-\frac23\int t^{1/2}dt=\frac23t^{3/2}\log t-\frac49t^{3/2} \]

Step 4: Final Form:
\[ I=-\frac12\left[\frac23t^{3/2}\log t-\frac49t^{3/2}\right]+c=t^{3/2}\left[-\frac13\log t+\frac29\right]+c \]
This is option (A). Options (C) and (D) have the constant \(2/3\), which would need \(\tfrac43\) in the bracket before halving, and (B) has the wrong sign on the constant.

Final Answer:
The integral is option (A). \[ \boxed{\text{(A)}} \]