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

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

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

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

x^{2} - arctan (x) + C

Solution steps

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

\int 2 x d x = x^{2}

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

= x^{2} - arctan (x)

Add a constant to the solution

= x^{2} - arctan (x) + C

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