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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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 = x + 3

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

Add a constant to the solution

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

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