What is the integral of (-3)/(4+x^2) ?

The integral of (-3)/(4+x^2) is -3/2 arctan(x/2)+C

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integral of (-3)/(4+x^2)

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

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

Solution steps

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

Take the constant out:

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

= - 3 \cdot \int \frac{1}{4 + x ^{2}} d x

Apply Integral Substitution

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

Use the common integral:

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

= - 3 \cdot \frac{1}{2} arctan (u)

Substitute back

u = \frac{x}{2}

= - 3 \cdot \frac{1}{2} arctan (\frac{x}{2})

Simplify

- 3 \cdot \frac{1}{2} arctan (\frac{x}{2}) : - \frac{3}{2} arctan (\frac{x}{2})

= - \frac{3}{2} arctan (\frac{x}{2})

Add a constant to the solution

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

What is the integral of (-3)/(4+x^2) ?
The integral of (-3)/(4+x^2) is -3/2 arctan(x/2)+C