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

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

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

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

- ln 2 - x^{2} + C

Solution steps

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

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

Simplify

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

= - ln 2 - x^{2}

Add a constant to the solution

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

\frac{d}{d x} (c + x + f (x) (x)^{2})

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

y = 3 (x^{2} - x + 1) (5 x - 3)

\frac{\partial}{\partial x} ((x + y^{2})^{3})

\frac{4}{x}, at x = - 5

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