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

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

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

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

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

Solution steps

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

Expand

\frac{- x + 1}{x ^{2} + 1} : - \frac{x}{x ^{2} + 1} + \frac{1}{x ^{2} + 1}

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

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

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

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

Add a constant to the solution

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

(\frac{1}{3})^{x}

\int_{0}^{1} 2 x + 1 d x

x \to \infty + lim (25 x^{4} + 9 x - 5 x^{2})

\frac{d}{d x} (2^{e})

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

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