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: Verify by Differentiating:
Let $F=t^{3/2}\left[-\tfrac13\log t+\tfrac29\right]$ with $t=1+x^{-2}$. Then $\dfrac{dF}{dt}=\tfrac32t^{1/2}\left[-\tfrac13\log t+\tfrac29\right]+t^{3/2}\left(-\tfrac1{3t}\right)$.

Step 2: Simplify:
$\dfrac{dF}{dt}=-\tfrac12t^{1/2}\log t+\tfrac13t^{1/2}-\tfrac13t^{1/2}=-\tfrac12t^{1/2}\log t$.

Step 3: Chain Rule:
$\dfrac{dt}{dx}=-2x^{-3}$, so $\dfrac{dF}{dx}=t^{1/2}\log t\cdot x^{-3}=\dfrac{\sqrt{x^2+1}}{x}\cdot\dfrac{\log t}{x^3}$, which is the given integrand. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A)}} \]