What is the integral of (x^2+9x+18)/(x+3) ?

The integral of (x^2+9x+18)/(x+3) is ((x+3)^2)/2+3(x+3)+C

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integral of (x^2+9x+18)/(x+3)

\int \frac{x ^{2} + 9 x + 18}{x + 3} d x

\frac{( x + 3 ) ^{2}}{2} + 3 (x + 3) + C

Solution steps

\int \frac{x ^{2} + 9 x + 18}{x + 3} d x

Apply u-substitution

= \int u + 3 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 3 d u

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

\int 3 d u = 3 u

= \frac{u ^{2}}{2} + 3 u

Substitute back

u = x + 3

= \frac{( x + 3 ) ^{2}}{2} + 3 (x + 3)

Add a constant to the solution

= \frac{( x + 3 ) ^{2}}{2} + 3 (x + 3) + C

What is the integral of (x^2+9x+18)/(x+3) ?
The integral of (x^2+9x+18)/(x+3) is ((x+3)^2)/2+3(x+3)+C