What is the integral of (x^2+5x+7)/(x+3) ?

The integral of (x^2+5x+7)/(x+3) is 1/2 (x+3)^2-x-3+ln|x+3|+C

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integral of (x^2+5x+7)/(x+3)

\int \frac{x ^{2} + 5 x + 7}{x + 3} d x

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

Solution steps

\int \frac{x ^{2} + 5 x + 7}{x + 3} d x

Apply u-substitution

: \int \frac{u ^{2} - u + 1}{u} d u

= \int \frac{u ^{2} - u + 1}{u} d u

Expand

\frac{u ^{2} - u + 1}{u} : u - 1 + \frac{1}{u}

= \int u - 1 + \frac{1}{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 1 d u + \int \frac{1}{u} d u

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - u + ln ∣ u ∣

Substitute back

u = x + 3

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

Simplify

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

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

Add a constant to the solution

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

\int y e^{4 y} d y

\frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

\frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

\frac{2 s ^{2} + 1}{s + 3}

What is the integral of (x^2+5x+7)/(x+3) ?
The integral of (x^2+5x+7)/(x+3) is 1/2 (x+3)^2-x-3+ln|x+3|+C