What is the integral from 0 to 2 of e^x-1/(x^2+1) ?

The integral from 0 to 2 of e^x-1/(x^2+1) is e^2-1-arctan(2)

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integral from 0 to 2 of e^x-1/(x^2+1)

\int_{0}^{2} e^{x} - \frac{1}{x ^{2} + 1} d x

e^{2} - 1 - arctan (2)

Decimal

5.28190 \dots

Solution steps

\int_{0}^{2} e^{x} - \frac{1}{x ^{2} + 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2} e^{x} d x - \int_{0}^{2} \frac{1}{x ^{2} + 1} d x

\int_{0}^{2} e^{x} d x = e^{2} - 1

\int_{0}^{2} \frac{1}{x ^{2} + 1} d x = arctan (2)

= e^{2} - 1 - arctan (2)

What is the integral from 0 to 2 of e^x-1/(x^2+1) ?
The integral from 0 to 2 of e^x-1/(x^2+1) is e^2-1-arctan(2)