What is the integral of (3x^2+14x+13)/(x+4) ?

The integral of (3x^2+14x+13)/(x+4) is -10x+5ln|x+4|+3/2 (x+4)^2-40+C

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integral of (3x^2+14x+13)/(x+4)

\int \frac{3 x ^{2} + 14 x + 13}{x + 4} d x

- 10 x + 5 ln ∣ x + 4 ∣ + \frac{3}{2} (x + 4)^{2} - 40 + C

Solution steps

\int \frac{3 x ^{2} + 14 x + 13}{x + 4} d x

Expand

\frac{3 x ^{2} + 14 x + 13}{x + 4} : \frac{3 x ^{2}}{x + 4} + \frac{14 x}{x + 4} + \frac{13}{x + 4}

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{3 x ^{2}}{x + 4} d x + \int \frac{14 x}{x + 4} d x + \int \frac{13}{x + 4} d x

\int \frac{3 x ^{2}}{x + 4} d x = 3 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣)

\int \frac{14 x}{x + 4} d x = 14 (x + 4 - 4 ln ∣ x + 4 ∣)

\int \frac{13}{x + 4} d x = 13 ln ∣ x + 4 ∣

= 3 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣) + 14 (x + 4 - 4 ln ∣ x + 4 ∣) + 13 ln ∣ x + 4 ∣

Simplify

3 (\frac{( x + 4 ) ^{2}}{2} - 8 (x + 4) + 16 ln ∣ x + 4 ∣) + 14 (x + 4 - 4 ln ∣ x + 4 ∣) + 13 ln ∣ x + 4 ∣ : - 10 x + 5 ln ∣ x + 4 ∣ + \frac{3}{2} (x + 4)^{2} - 40

= - 10 x + 5 ln ∣ x + 4 ∣ + \frac{3}{2} (x + 4)^{2} - 40

Add a constant to the solution

= - 10 x + 5 ln ∣ x + 4 ∣ + \frac{3}{2} (x + 4)^{2} - 40 + C

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

\int \frac{x ^{2} - 14 x - 5}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

f (x) = 9 x + 5

y^{^{'}} = t^{2}

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

What is the integral of (3x^2+14x+13)/(x+4) ?
The integral of (3x^2+14x+13)/(x+4) is -10x+5ln|x+4|+3/2 (x+4)^2-40+C