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

The integral of x/(4-x^2) is -1/2 ln|4-x^2|+C

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integral of x/(4-x^2)

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

- \frac{1}{2} ln 4 - x^{2} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 4 - x^{2}

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

Add a constant to the solution

= - \frac{1}{2} ln 4 - x^{2} + C

\frac{\partial}{\partial x} (11 x^{2} - 2 x^{3})

\int_{0}^{1} 2 π (x) (9 - 8 x - x^{2}) d x

g (x) = \frac{1 + 6 x}{9 - 4 x}

f^{^{'}} (x) = 3^{x}

\int + (8 x + 3 sin (x))^{2} d x

What is the integral of x/(4-x^2) ?
The integral of x/(4-x^2) is -1/2 ln|4-x^2|+C