Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन2 Marks
Question
ज्ञात कीजिए : $\int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x$
✓
Answer
मान लीजिए कि $\mathrm{I}=\int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x $
$=\int e^{x}\left[\frac{\left.x^{2}-1+1+1\right)}{(x+1)^{2}}\right] dx$
$=\int e^{x}\left[\frac{x^{2}-1}{(x+1)^{2}}+\frac{2}{(x+1)^{2}}\right] d x $
$=\int e^{x}\left[\frac{x-1}{x+1}+\frac{2}{(x+1)^{2}}\right] d x$
मान लीजिए कि $f (x) = \frac{x-1}{x+1} तब f'(x) = \frac{2}{(x+1)^{2}}$
अतः दिया हुआ समाकल्य $e^x [f(x) + f′(x)]$ के रूप में है।
इसलिए $\int \frac{x^{2}+1}{(x+1)^{2}} e^{x} d x =\frac{x-1}{x+1} e^{x}+\mathrm{C}$
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