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

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

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

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

x - 3 arctan (\frac{x}{3}) + C

Solution steps

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

Apply Integral Substitution

= \int 3 (1 - \frac{1}{u ^{2} + 1}) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int 1 - \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int 1 d u - \int \frac{1}{u ^{2} + 1} d u)

\int 1 d u = u

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

= 3 (u - arctan (u))

Substitute back

u = \frac{x}{3}

= 3 (\frac{x}{3} - arctan (\frac{x}{3}))

Simplify

3 (\frac{x}{3} - arctan (\frac{x}{3})) : x - 3 arctan (\frac{x}{3})

= x - 3 arctan (\frac{x}{3})

Add a constant to the solution

= x - 3 arctan (\frac{x}{3}) + C

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