What is the integral of ((x^2+2x+1))/((x+3)) ?

The integral of ((x^2+2x+1))/((x+3)) is -4x+4ln|x+3|+1/2 (x+3)^2-12+C

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integral of ((x^2+2x+1))/((x+3))

\int \frac{( x ^{2} + 2 x + 1 )}{( x + 3 )} d x

- 4 x + 4 ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - 12 + C

Solution steps

\int \frac{x ^{2} + 2 x + 1}{x + 3} d x

Expand

\frac{x ^{2} + 2 x + 1}{x + 3} : \frac{x ^{2}}{x + 3} + \frac{2 x}{x + 3} + \frac{1}{x + 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{2}}{x + 3} d x + \int \frac{2 x}{x + 3} d x + \int \frac{1}{x + 3} d x

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

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

\int \frac{1}{x + 3} d x = ln ∣ x + 3 ∣

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

Simplify

\frac{( x + 3 ) ^{2}}{2} - 6 (x + 3) + 9 ln ∣ x + 3 ∣ + 2 (x + 3 - 3 ln ∣ x + 3 ∣) + ln ∣ x + 3 ∣ : - 4 x + 4 ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - 12

= - 4 x + 4 ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - 12

Add a constant to the solution

= - 4 x + 4 ln ∣ x + 3 ∣ + \frac{1}{2} (x + 3)^{2} - 12 + C

\frac{d y}{d t} = y^{2} - 4

f (x) = 7 x + 1

y^{^{'}} = \frac{x}{1 + 2 y}, y (- 1) = 0

f (x) = \frac{π}{3}

\int \frac{3}{( 2 x - 1 )} d x

What is the integral of ((x^2+2x+1))/((x+3)) ?
The integral of ((x^2+2x+1))/((x+3)) is -4x+4ln|x+3|+1/2 (x+3)^2-12+C