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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

x \to 2 - lim (8 x - x^{4})

\int \frac{e ^{\frac{- 1}{x}}}{x ^{2}} d x

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

\frac{d}{d x} (ln (3 x) - e^{s e c (x)})

sin (tan (8 x))

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