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

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

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

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

x - ln x^{2} + 1 + C

Solution steps

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

\int 1 d x = x

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

= x - ln x^{2} + 1

Add a constant to the solution

= x - ln x^{2} + 1 + C

\int \frac{x ^{3} - 2 x ^{4} + x ^{5}}{2} d x

3 tan (x)

f (x) = 12 x, (9, 36)

n = 0 \sum \infty \frac{1}{2} \cdot (- 1)^{n}

\frac{\partial}{\partial x} (- 4 x y - x^{4} - y^{4})

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