Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
$$\int \frac{dx}{\prod_{i=0}^{n}(x+r)} \text{ is equal to:}$$
$$\frac{1}{n!}\left[\sum_{r=0}^{n}(-1)^r \cdot ^nC_r \ln(x+r)\right] + c$$
$$\frac{1}{n!}\left[\sum_{r=0}^{n-1}(-1)^r \cdot \ln(x+r-1)\right] + c$$
$$\frac{1}{n!}\ln\left(\prod_{r=0}^{n}(x+r)^{(-1)^r \cdot ^nC_r}\right) + c$$
$$\frac{1}{n!}\sum_{r=0}^{n-1}\ln(x+r-1)^{(-1)^{r-1}}$$
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