What is the integral of (x^2+5x+6)/(x+3) ?

The integral of (x^2+5x+6)/(x+3) is 1/2 (x+3)^2-x-3+C

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integral of (x^2+5x+6)/(x+3)

\int \frac{x ^{2} + 5 x + 6}{x + 3} d x

\frac{1}{2} (x + 3)^{2} - x - 3 + C

Solution steps

\int \frac{x ^{2} + 5 x + 6}{x + 3} d x

Apply u-substitution

: \int u - 1 d u

= \int u - 1 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 1 d u

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

\int 1 d u = u

= \frac{u ^{2}}{2} - u

Substitute back

u = x + 3

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

Simplify

\frac{( x + 3 ) ^{2}}{2} - (x + 3) : \frac{1}{2} (x + 3)^{2} - x - 3

= \frac{1}{2} (x + 3)^{2} - x - 3

Add a constant to the solution

= \frac{1}{2} (x + 3)^{2} - x - 3 + C

y = 9 - x^{2}, y = x + 3

\int sin (6 x) d x

(- 72 sin (t))^{^{'}}

y = 2 x^{3}

\frac{d}{d x} (f (x) (x)^{n})

What is the integral of (x^2+5x+6)/(x+3) ?
The integral of (x^2+5x+6)/(x+3) is 1/2 (x+3)^2-x-3+C