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

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

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

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

- ln 4 - x^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 4 - x^{2}

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

Simplify

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

= - ln 4 - x^{2}

Add a constant to the solution

= - ln 4 - x^{2} + C

y^{^{''}} + 2 y = - 18 x^{2} e^{2} x

\frac{d}{d x} (- \frac{9}{( 1 + 3 x ) ^{2}})

y = x^{3} - 3 x^{2} + 3 x, y = x

\frac{d}{d x} (x^{2} - 7 x + 12)

\frac{d}{d x} (\frac{e ^{\frac{x}{2}}}{x + 1})

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