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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\int \frac{x ^{2}}{x + 3} d x

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

y = 11 x, y = x^{2} - 12

\int_{1}^{4} (x^{2}) d x

\int_{1}^{3} 1 + \frac{1}{x ^{2}} d x

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