What is the integral of (-x-1)/(x^2+3) ?

The integral of (-x-1)/(x^2+3) is -1/2 ln|x^2+3|-1/(sqrt(3))arctan(x/(sqrt(3)))+C

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integral of (-x-1)/(x^2+3)

\int \frac{- x - 1}{x ^{2} + 3} d x

- \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{- x - 1}{x ^{2} + 3} d x

Expand

\frac{- x - 1}{x ^{2} + 3} : - \frac{x}{x ^{2} + 3} - \frac{1}{x ^{2} + 3}

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} + 3} d x - \int \frac{1}{x ^{2} + 3} d x

\int \frac{x}{x ^{2} + 3} d x = \frac{1}{2} ln x^{2} + 3

\int \frac{1}{x ^{2} + 3} d x = \frac{1}{3} arctan (\frac{x}{3})

= - \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= - \frac{1}{2} ln x^{2} + 3 - \frac{1}{3} arctan (\frac{x}{3}) + C

\int x x^{\frac{3}{2}} - 1 d x

sin (x), cos (x), x = - \frac{π}{6}

\frac{d}{d x} (x^{2} e^{5 x})

7 + 4 x

\int_{0}^{4} - x^{2} d x

What is the integral of (-x-1)/(x^2+3) ?
The integral of (-x-1)/(x^2+3) is -1/2 ln|x^2+3|-1/(sqrt(3))arctan(x/(sqrt(3)))+C