यहाँ $\mathrm{I}=\int e^{x}\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right)$dx
अब $f(x) = \tan^{-1} x$, लीजिए, तब f'(x) = $\frac{1}{1+x^{2}}$
अतः दिया हुआ समाकल्य $e^x[f(x) + f′(x)]$ के रूप में है।
इसलिए $\mathrm{I}=\int e^{x}$$\left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) dx = e^xtan^{− 1}x + C$
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