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

y d y + (1 + y^{2}) sin (t) d t = 0

\frac{d}{d x} (e x)

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

y = 1 + x

y = x, y = 2 x, x = 1

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