Question

$\int \frac{x^4 \cos(\tan^{-1} x^5)}{1 + x^{10}} dx$ equals ______.

• A. $\sin(\tan^{-1} x^5) + c$
• B. $x^4 \sin(\tan^{-1} x^5) + c$
• C. $\frac{1}{5}\sin(\tan^{-1} x^5) + c$
• D. $\cos(\tan^{-1} x^5) + c$

Hint:

Always look for the function-derivative pair in integration problems. If you see $f(g(x))$ multiplied by something that strongly resembles $g'(x)$, substitution $t = g(x)$ will almost certainly solve it.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use the substitution method. Notice that the derivative of $\tan^{-1}(x^5)$ involves $x^4$ and $1 + x^{10}$.

Step 2: Formula Application:
Let $t = \tan^{-1}(x^5)$. Then $dt = \frac{1}{1 + (x^5)^2} \cdot 5x^4 \, dx = \frac{5x^4}{1 + x^{10}} \, dx$. $\implies \frac{1}{5} dt = \frac{x^4}{1 + x^{10}} \, dx$.

Step 3: Explanation:
The integral becomes: $\int \cos(t) \cdot \frac{1}{5} \, dt = \frac{1}{5} \sin(t) + c$. Substituting $t$ back: $\frac{1}{5} \sin(\tan^{-1} x^5) + c$.

Step 4: Final Answer:
The integral equals $\frac{1}{5}\sin(\tan^{-1} x^5) + c$. (Note: Option C in the original list is usually corrected to include the 1/5 factor).