What is the integral of (x^2)/(9+x) ?

The integral of (x^2)/(9+x) is ((9+x)^2)/2-18(9+x)+81ln|9+x|+C

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integral of (x^2)/(9+x)

\int \frac{x ^{2}}{9 + x} d x

\frac{( 9 + x ) ^{2}}{2} - 18 (9 + x) + 81 ln ∣ 9 + x ∣ + C

Solution steps

\int \frac{x ^{2}}{9 + x} d x

Apply u-substitution

: \int \frac{( u - 9 ) ^{2}}{u} d u

= \int \frac{( u - 9 ) ^{2}}{u} d u

Expand

\frac{( u - 9 ) ^{2}}{u} : u - 18 + \frac{81}{u}

= \int u - 18 + \frac{81}{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 18 d u + \int \frac{81}{u} d u

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

\int 18 d u = 18 u

\int \frac{81}{u} d u = 81 ln ∣ u ∣

= \frac{u ^{2}}{2} - 18 u + 81 ln ∣ u ∣

Substitute back

u = 9 + x

= \frac{( 9 + x ) ^{2}}{2} - 18 (9 + x) + 81 ln ∣ 9 + x ∣

Add a constant to the solution

= \frac{( 9 + x ) ^{2}}{2} - 18 (9 + x) + 81 ln ∣ 9 + x ∣ + C

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What is the integral of (x^2)/(9+x) ?
The integral of (x^2)/(9+x) is ((9+x)^2)/2-18(9+x)+81ln|9+x|+C