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

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

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integral of 3/(x^2+4)

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

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

Solution steps

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

\int x^{4} x^{5} + 3 d x

\frac{\partial}{\partial y} (2 e^{\frac{x}{2} + y} - y)

\frac{490 t}{( t ^{2} + 7 ) ^{3}}

\frac{d}{d x} (\frac{x ^{2} ( x + 2 ) ^{4}}{( 2 x ^{2} - 1 ) ^{3}})

sin (t - 8)

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