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 $\mathrm{I}=\int e^{x} \left(\tan ^{-1} x+\frac{1}{1+x^{2}}\right) dx = e^{x }\tan^{− 1}x + C$
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