Same as that of $\int_0^{\pi} \frac{x^2+1}{1+x^4} dx$
$\frac{\pi}{2\sqrt{2}}$
Same as that of $\int_0^{\pi} \frac{x^2 dx}{1+x^4}$
$\frac{\pi}{\sqrt{2}}$
Step-by-Step Solution
Key Concept: Recognize the binomial coefficient structure in partial fractions and use logarithm properties to combine the result into a product form.
For the integral $\int \frac{dx}{x(x+1)(x+2)\cdots(x+n)}$, use partial fraction decomposition with coefficients $^nC_r$ defined recursively. The decomposition is $\frac{1}{x(x+1)\cdots(x+n)} = \frac{1}{n!}\left[\frac{^nC_0}{x} + \frac{^nC_1}{x+1} + \cdots + \frac{(-1)^n\,^nC_n}{x+n}\right]$. Integration gives $I = \frac{1}{n!}\left[^nC_0\ln x - ^nC_1\ln(x+1) + \cdots + (-1)^n\,^nC_n\ln(x+n)\right] + c = \frac{1}{n!}\ln\left[\prod_{r=0}^{n}(x+r)^{(-1)^r\,^nC_r}\right] + c$.
Correct Answer: 1,3