What is the integral of (-x+1)/(x^2+4) ?

The integral of (-x+1)/(x^2+4) is -1/2 ln|x^2+4|+1/2 arctan(x/2)+C

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integral of (-x+1)/(x^2+4)

\int \frac{- x + 1}{x ^{2} + 4} d x

- \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{- x + 1}{x ^{2} + 4} d x

Expand

\frac{- x + 1}{x ^{2} + 4} : - \frac{x}{x ^{2} + 4} + \frac{1}{x ^{2} + 4}

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} + 4} d x + \int \frac{1}{x ^{2} + 4} d x

\int \frac{x}{x ^{2} + 4} d x = \frac{1}{2} ln x^{2} + 4

\int \frac{1}{x ^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2})

= - \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2})

Add a constant to the solution

= - \frac{1}{2} ln x^{2} + 4 + \frac{1}{2} arctan (\frac{x}{2}) + C

\int_{2}^{\infty} e^{- 4 p} d p

\int \frac{cos ( x )}{7 - sin ( x )} d x

\frac{\partial}{\partial x} (x^{6} y - 3 x^{3} y^{2})

\int \frac{1}{9 x ^{2} + 4} d x

(x - 6)^{\frac{8}{9}}

What is the integral of (-x+1)/(x^2+4) ?
The integral of (-x+1)/(x^2+4) is -1/2 ln|x^2+4|+1/2 arctan(x/2)+C