What is the integral of (1+x)/(1+x^2) ?

The integral of (1+x)/(1+x^2) is arctan(x)+1/2 ln|1+x^2|+C

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integral of (1+x)/(1+x^2)

\int \frac{1 + x}{1 + x ^{2}} d x

arctan (x) + \frac{1}{2} ln 1 + x^{2} + C

Solution steps

\int \frac{1 + x}{1 + x ^{2}} d x

Expand

\frac{1 + x}{1 + x ^{2}} : \frac{1}{1 + x ^{2}} + \frac{x}{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{1}{1 + x ^{2}} d x + \int \frac{x}{1 + x ^{2}} d x

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

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

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

Add a constant to the solution

= arctan (x) + \frac{1}{2} ln 1 + x^{2} + C

x \to 3 - lim (\frac{( 2 )}{x - 3})

x \to - \infty lim (e^{- 6 x} - 2 e^{- 2 x} - 1)

\int \frac{x ^{2}}{( 1 - x ^{3} ) ^{\frac{3}{2}}} d x

\int_{0}^{4} π (4 x^{2})^{2} d x

\frac{\partial}{\partial x} (5 + 2 x + 3 x y^{2})

What is the integral of (1+x)/(1+x^2) ?
The integral of (1+x)/(1+x^2) is arctan(x)+1/2 ln|1+x^2|+C