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

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

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

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

- \frac{3}{2} ln 4 - x^{2} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 4 - x^{2}

= 3 (- \frac{1}{2} ln 4 - x^{2})

Simplify

3 (- \frac{1}{2} ln 4 - x^{2}) : - \frac{3}{2} ln 4 - x^{2}

= - \frac{3}{2} ln 4 - x^{2}

Add a constant to the solution

= - \frac{3}{2} ln 4 - x^{2} + C

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

\int a rcco tan (2 x) d x

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

sin (x) 1 - cos (x)

\int \frac{3}{x ^{2} - 9} d x

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