Integral Calculus
86386
If \(I_{m, n}=\int_{0}^{1} x^{m}(\log x)^{n} d x\), then it is equal to
1 \(\frac{n}{n+1} I_{m, n-1}\)
2 \(\frac{-m}{n+1} I_{m, n-1}\)
3 \(\frac{-n}{m+1} I_{m, n-1}\)
4 \(\frac{\mathrm{m}}{\mathrm{n}+1} \mathrm{I}_{\mathrm{m}, \mathrm{n}-1}\)
Explanation:
(C): Given,
\(I_{m, n}=\int_{0}^{1} x^{m}(\log x)^{n} d x\)
\(=\left[(\log x)^{n} \cdot \frac{x^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1}\left(n(\log x)^{n-1} \cdot \frac{1}{x} \cdot \frac{x^{m+1}}{m+1}\right) d x\)
\(=0-\frac{\mathrm{n}}{\mathrm{m}+1} \int_{0}^{1} \mathrm{x}^{\mathrm{m}}(\log \mathrm{x})^{\mathrm{n}-1} \mathrm{dx}\)
\(=\frac{-\mathrm{n}}{\mathrm{m}+1} \cdot \mathrm{I}_{\mathrm{m}, \mathrm{n}-1}\)