Question

\(\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1}\,dx = \cdots\)

• A. \(tan^{-1}(x^5+\frac{1}{x^5})+c\)
• B. \(\frac{1}{5}tan^{-1}(x^5+\frac{1}{x^5})+c\)
• C. \(tan^{-1}(x^{10}+\frac{1}{x^{10}})+c\)
• D. \(\frac{1}{10}tan^{-1}(x^{10}+\frac{1}{x^{10}})+c\)

Hint:

Divide top and bottom by x^10 and substitute t = x^5 + 1/x^5.

Answer 1

The Correct Option is B

Step 1: Check by differentiating option (B):
If $F = \frac15\tan^{-1}t$ with $t = x^5 + x^{-5}$, then $F' = \frac15\cdot\frac{t'}{1+t^2}$.

Step 2: Compute:
$t' = 5x^4 - 5x^{-6} = 5(x^{10}-1)x^{-6}$. And $1 + t^2 = x^{10} + 3 + x^{-10} = \frac{x^{20}+3x^{10}+1}{x^{10}}$.

Step 3: Combine:
$F' = \frac15\cdot\frac{5(x^{10}-1)x^{-6}\,x^{10}}{x^{20}+3x^{10}+1} = \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1}$, which is exactly the integrand.

Final Answer:
Option (B) is the integral. \[ \boxed{\frac{1}{5}\tan^{-1}\left(x^5+\frac{1}{x^5}\right)+c} \]