(B) : Given, \(\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) We know that, \(\int e^{x}\left[\int(x)+\int^{\prime}(x)\right] d x=e^{x} \int(x)+c\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) \(=e^{x} \frac{1}{x}+c=\frac{e^{x}}{x}+c\)
(B) : Given, \(\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) We know that, \(\int e^{x}\left[\int(x)+\int^{\prime}(x)\right] d x=e^{x} \int(x)+c\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) \(=e^{x} \frac{1}{x}+c=\frac{e^{x}}{x}+c\)
(B) : Given, \(\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) We know that, \(\int e^{x}\left[\int(x)+\int^{\prime}(x)\right] d x=e^{x} \int(x)+c\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) \(=e^{x} \frac{1}{x}+c=\frac{e^{x}}{x}+c\)
(B) : Given, \(\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) We know that, \(\int e^{x}\left[\int(x)+\int^{\prime}(x)\right] d x=e^{x} \int(x)+c\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) \(=e^{x} \frac{1}{x}+c=\frac{e^{x}}{x}+c\)
(B) : Given, \(\int\left(\frac{x-1}{x^{2}}\right) e^{x} d x\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) We know that, \(\int e^{x}\left[\int(x)+\int^{\prime}(x)\right] d x=e^{x} \int(x)+c\) \(\int\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x\) \(=e^{x} \frac{1}{x}+c=\frac{e^{x}}{x}+c\)