The value of int frac{e^{x}left(x^{2} tan ^{ | vedclass

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Question

(A) Given that,$ I = \int \frac{e^{x}(x^{2} 	an^{-1} x + 	an^{-1} x + 1)}{x^{2} + 1} dx $.We can rewrite the numerator by grouping terms:$ I = \int

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$ e^{x} 	an ^{-1} x+c $

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(A) Given that,$ I = \int \frac{e^{x}(x^{2} 	an^{-1} x + 	an^{-1} x + 1)}{x^{2} + 1} dx $.We can rewrite the numerator by grouping terms:$ I = \int e^{x} \left( \frac{(x^{2} + 1) 	an^{-1} x + 1}{x^{2} + 1} ight) dx $.Dividing each term by $ x^{2} + 1 $:$ I = \int e^{x} \left( 	an^{-1} x + \frac{1}{x^{2} + 1} ight) dx $.Let $ f(x) = 	an^{-1} x $. Then $ f'(x) = \frac{1}{x^{2} + 1} $.Using the standard integral formula $ \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c $:$ I = e^{x} 	an^{-1} x + c $.

The value of $ \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $ is equal to

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