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

The integral of (4-x^2)/(2+x) is 4x-1/2 (x+2)^2+8+C

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

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

4 x - \frac{1}{2} (x + 2)^{2} + 8 + C

Solution steps

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

Expand

\frac{4 - x ^{2}}{2 + x} : \frac{4}{2 + x} - \frac{x ^{2}}{2 + x}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{4}{2 + x} d x = 4 ln ∣ 2 + x ∣

\int \frac{x ^{2}}{2 + x} d x = \frac{( 2 + x ) ^{2}}{2} - 4 (2 + x) + 4 ln ∣ 2 + x ∣

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

Simplify

4 ln ∣ 2 + x ∣ - (\frac{( 2 + x ) ^{2}}{2} - 4 (2 + x) + 4 ln ∣ 2 + x ∣) : 4 x - \frac{1}{2} (x + 2)^{2} + 8

= 4 x - \frac{1}{2} (x + 2)^{2} + 8

Add a constant to the solution

= 4 x - \frac{1}{2} (x + 2)^{2} + 8 + C

What is the integral of (4-x^2)/(2+x) ?
The integral of (4-x^2)/(2+x) is 4x-1/2 (x+2)^2+8+C