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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{( u - 2 ) ^{2}}{u} d u

Expand

\frac{( u - 2 ) ^{2}}{u} : u - 4 + \frac{4}{u}

= \int u - 4 + \frac{4}{u} d u

Apply the Sum Rule:

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

= \int u d u - \int 4 d u + \int \frac{4}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 4 d u = 4 u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{u ^{2}}{2} - 4 u + 4 ln ∣ u ∣

Substitute back

u = x + 2

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

Add a constant to the solution

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

cos (2 x + y)

\frac{1}{n ^{2}}

3^{2} x

\frac{( x ^{3} - 3 x )}{( 1 - x ^{4} )}

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

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