Question

If $I_{n} = \int \frac{t^{n}}{1+t^{2}} dt$ then $I_{6} + I_{4} =$

• A. $t^{3}/3$
• B. $t^{4}/4$
• C. $t^{5}/5$
• D. $t^{7}/7$

Hint:

When you see $I_{n} + I_{n-2}$, look for factors that cancel the denominator.

Answer 1

The Correct Option is C

Step 1: Concept
Combine the integrals into a single expression under one integral sign.

Step 2: Meaning
$I_{6} + I_{4} = \int \frac{t^{6}}{1+t^{2}} dt + \int \frac{t^{4}}{1+t^{2}} dt = \int \frac{t^{6} + t^{4}}{1+t^{2}} dt$.

Step 3: Analysis
Factor the numerator: $t^{6} + t^{4} = t^{4}(t^{2} + 1)$. The expression becomes $\int \frac{t^{4}(t^{2}+1)}{1+t^{2}} dt = \int t^{4} dt$.

Step 4: Conclusion
$\int t^{4} dt = \frac{t^{5}}{5}$. Thus $I_{6} + I_{4} = t^{5}/5$. Final Answer: (C)