What is the integral of (5x^2+20x+6)/(x^3+2x^2+x) ?

The integral of (5x^2+20x+6)/(x^3+2x^2+x) is 6ln|x|-ln|x+1|-9/(x+1)+C

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integral of (5x^2+20x+6)/(x^3+2x^2+x)

\int \frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} d x

6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + C

Solution steps

\int \frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} d x

Take the partial fraction of

\frac{5 x ^{2} + 20 x + 6}{x ^{3} + 2 x ^{2} + x} : \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{( x + 1 ) ^{2}}

= \int \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{( x + 1 ) ^{2}} d x

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{6}{x} d x - \int \frac{1}{x + 1} d x + \int \frac{9}{( x + 1 ) ^{2}} d x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

\int \frac{9}{( x + 1 ) ^{2}} d x = - \frac{9}{x + 1}

= 6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1}

Add a constant to the solution

= 6 ln ∣ x ∣ - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + C

\int \frac{2}{x + 5} - cos (3 x) d x

\frac{8 x - 6}{x - 6}

y^{^{'}} - 4 y = 0.8

(x - 1)^{2} (x - 2) (x - 4)

\int sinh (3 u) d u

What is the integral of (5x^2+20x+6)/(x^3+2x^2+x) ?
The integral of (5x^2+20x+6)/(x^3+2x^2+x) is 6ln|x|-ln|x+1|-9/(x+1)+C