What is the integral of (x^2)/(3+x) ?

The integral of (x^2)/(3+x) is ((3+x)^2)/2-6(3+x)+9ln|3+x|+C

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integral of (x^2)/(3+x)

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

\frac{( 3 + x ) ^{2}}{2} - 6 (3 + x) + 9 ln ∣ 3 + x ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{( u - 3 ) ^{2}}{u} d u

= \int \frac{( u - 3 ) ^{2}}{u} d u

Expand

\frac{( u - 3 ) ^{2}}{u} : u - 6 + \frac{9}{u}

= \int u - 6 + \frac{9}{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 6 d u + \int \frac{9}{u} d u

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

\int 6 d u = 6 u

\int \frac{9}{u} d u = 9 ln ∣ u ∣

= \frac{u ^{2}}{2} - 6 u + 9 ln ∣ u ∣

Substitute back

u = 3 + x

= \frac{( 3 + x ) ^{2}}{2} - 6 (3 + x) + 9 ln ∣ 3 + x ∣

Add a constant to the solution

= \frac{( 3 + x ) ^{2}}{2} - 6 (3 + x) + 9 ln ∣ 3 + x ∣ + C

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What is the integral of (x^2)/(3+x) ?
The integral of (x^2)/(3+x) is ((3+x)^2)/2-6(3+x)+9ln|3+x|+C