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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 ( u + u )} d u

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

Apply u-substitution

: \int \frac{2}{v + 1} d v

= - \frac{1}{2} \cdot \int \frac{2}{v + 1} d v

Take the constant out:

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

= - \frac{1}{2} \cdot 2 \cdot \int \frac{1}{v + 1} d v

Apply u-substitution

: \int \frac{1}{w} d w

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

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= - \frac{1}{2} \cdot 2 ln ∣ w ∣

Substitute back

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

Simplify

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

= - ln 4 - x^{2} + 1

Add a constant to the solution

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

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