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

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

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integral of (e^{arctan(x)}+1)/(1+x^2)

\int \frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} d x

e^{a r c t a n (x)} + arctan (x) + C

Solution steps

\int \frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} d x

Expand

\frac{e ^{a r c t a n (x)} + 1}{1 + x ^{2}} : \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} + \frac{1}{1 + x ^{2}}

Apply the Sum Rule:

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

= \int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x + \int \frac{1}{1 + x ^{2}} d x

\int \frac{e ^{a r c t a n (x)}}{1 + x ^{2}} d x = e^{a r c t a n (x)}

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

= e^{a r c t a n (x)} + arctan (x)

Add a constant to the solution

= e^{a r c t a n (x)} + arctan (x) + C

x \to - 1 - lim (\frac{x}{∣ x - 1 ∣})

\int \frac{1}{10 r ^{\frac{1}{2}} - r} d r

\frac{d}{d x} (15 x^{2} + 2 x - 8)

\int 4 sin (x + \frac{π}{4}) cos (x - \frac{π}{4}) d x

\frac{\partial}{\partial x} (\frac{x ( v - b )}{r ( x , v )})

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