What is the integral of e^x+20(1+x^2)^{-1} ?

The integral of e^x+20(1+x^2)^{-1} is e^x+20arctan(x)+C

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integral of e^x+20(1+x^2)^{-1}

\int e^{x} + 20 (1 + x^{2})^{- 1} d x

e^{x} + 20 arctan (x) + C

Solution steps

\int e^{x} + 20 (1 + x^{2})^{- 1} d x

20 (1 + x^{2})^{- 1} = \frac{20}{1 + x ^{2}}

= \int e^{x} + \frac{20}{1 + x ^{2}} 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 e^{x} d x + \int \frac{20}{1 + x ^{2}} d x

\int e^{x} d x = e^{x}

\int \frac{20}{1 + x ^{2}} d x = 20 arctan (x)

= e^{x} + 20 arctan (x)

Add a constant to the solution

= e^{x} + 20 arctan (x) + C

\frac{d}{d x} (sin (\frac{x}{4}))

x \to \frac{x}{2} lim (sin (x))

\int - sin (x) \cdot cos (x) d x

\frac{d}{d x} (\frac{2}{( x ^{2} - 3 x ) ^{2}})

\frac{d}{d x} ((x^{2} + a^{2})^{5})

What is the integral of e^x+20(1+x^2)^{-1} ?
The integral of e^x+20(1+x^2)^{-1} is e^x+20arctan(x)+C