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

The integral of ((x^2-2))/((x+4)) is 1/2 (x+4)^2-8x-32+14ln|x+4|+C

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

integral

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

\frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{x + 4} d x = \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣

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

= \frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ - 2 ln ∣ x + 4 ∣

Simplify

\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣ - 2 ln ∣ x + 4 ∣ : \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣

= \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣

Add a constant to the solution

= \frac{1}{2} (x + 4)^{2} - 8 x - 32 + 14 ln ∣ x + 4 ∣ + C

in t e g r a l (\frac{1}{x}) sin (\frac{1}{x})

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in t e g r a l \frac{( 2 v + 5 )}{( 7 v + 22 )}

What is the integral of ((x^2-2))/((x+4)) ?
The integral of ((x^2-2))/((x+4)) is 1/2 (x+4)^2-8x-32+14ln|x+4|+C