What is the integral of (5x^2+20x+6)/(x(x+1)^2) ?

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

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integral of (5x^2+20x+6)/(x(x+1)^2)

\int \frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} d x

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

Solution steps

\int \frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} d x

Expand

\frac{5 x ^{2} + 20 x + 6}{x ( x + 1 ) ^{2}} : \frac{5 x ^{2}}{x ( x + 1 ) ^{2}} + \frac{20 x}{x ( x + 1 ) ^{2}} + \frac{6}{x ( x + 1 ) ^{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{5 x ^{2}}{x ( x + 1 ) ^{2}} d x + \int \frac{20 x}{x ( x + 1 ) ^{2}} d x + \int \frac{6}{x ( x + 1 ) ^{2}} d x

\int \frac{5 x ^{2}}{x ( x + 1 ) ^{2}} d x = 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1})

\int \frac{20 x}{x ( x + 1 ) ^{2}} d x = - \frac{20}{x + 1}

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

= 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{20}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Simplify

5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{20}{x + 1} + 6 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : - ln ∣ x + 1 ∣ - \frac{9}{x + 1} + 6 ln ∣ x ∣

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

Add a constant to the solution

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

\int x (2 x - 3) d x

\frac{\partial}{\partial x} (x^{3} - 6 x y + y^{3} - 2)

\frac{d}{d t} (+ 6 t)

\int_{4}^{10} \frac{1}{36 + ( x - 4 ) ^{2}} d x

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