Question

निम्नलिखित के मान निकालिए-

${\displaystyle \int \frac{{\text{e}}^{6\log x}-{\text{e}}^{5\log x}}{{\text{e}}^{4\log x}-{\text{e}}^{3\log x}}\text{d}x}$

Answer 1

माना I = ${\displaystyle \int \frac{{\text{e}}^{6\log x}-{\text{e}}^{5\log x}}{{\text{e}}^{4\log x}-{\text{e}}^{3\log x}}\text{d}x}$

= ${\displaystyle \int \frac{{\text{e}}^{{\log x}^6}-{\text{e}}^{{\log x}^5}}{{\text{e}}^{{\log x}^4}-{\text{e}}^{{\log x}^3}}\text{d}x}$  .....[∵ a log b – log b^a]

= ${\displaystyle \int \frac{x^6-x^5}{x^4-x^3}\text{d}x}$  .....[∵ e^logx = x]

= ${\displaystyle \int \frac{x^3-x^2}{x-1}\text{d}x}$

= ${\displaystyle \int \frac{x^2(x-1)}{x-1}\text{d}x}$

= ${\displaystyle \int x^2\text{d}x}$

= ${\displaystyle \frac{x^3}{3}+\text{C}}$