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

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

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

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

2 ln ∣ x + 4 ∣ - \frac{1}{x} + C

Solution steps

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

Expand

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

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

\int \frac{2 x ^{2}}{x ^{3} + 4 x ^{2}} d x = 2 ln ∣ x + 4 ∣

\int \frac{x}{x ^{3} + 4 x ^{2}} d x = \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣

\int \frac{4}{x ^{3} + 4 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

= 2 ln ∣ x + 4 ∣ + \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

Simplify

2 ln ∣ x + 4 ∣ + \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣ : 2 ln ∣ x + 4 ∣ - \frac{1}{x}

= 2 ln ∣ x + 4 ∣ - \frac{1}{x}

Add a constant to the solution

= 2 ln ∣ x + 4 ∣ - \frac{1}{x} + C

\int \frac{6}{sin ^{2} ( x )} d x

\frac{4 x}{x - 3}, at x = 5

f (x) = \frac{x ^{3}}{3} + x^{2} + 4 x - 10

y = 10 x - x^{2}, (5, 26)

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

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