What is the integral of (5x^2+5x+3)/(x(x+1)^2) ?

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

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integral of (5x^2+5x+3)/(x(x+1)^2)

\int \frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} d x

2 ln ∣ x + 1 ∣ + \frac{3}{x + 1} + 3 ln ∣ x ∣ + C

Solution steps

\int \frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} d x

Expand

\frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} : \frac{5 x ^{2}}{x ( x + 1 ) ^{2}} + \frac{5 x}{x ( x + 1 ) ^{2}} + \frac{3}{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{5 x}{x ( x + 1 ) ^{2}} d x + \int \frac{3}{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{5 x}{x ( x + 1 ) ^{2}} d x = - \frac{5}{x + 1}

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

= 5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1} + 3 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1})

Simplify

5 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1} + 3 (ln ∣ x ∣ - ln ∣ x + 1 ∣ + \frac{1}{x + 1}) : 2 ln ∣ x + 1 ∣ + \frac{3}{x + 1} + 3 ln ∣ x ∣

= 2 ln ∣ x + 1 ∣ + \frac{3}{x + 1} + 3 ln ∣ x ∣

Add a constant to the solution

= 2 ln ∣ x + 1 ∣ + \frac{3}{x + 1} + 3 ln ∣ x ∣ + C

\frac{d}{d x} (\frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 3

\int \frac{1}{1 - 3 y} d y

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)

f (x) = 5 x^{2} - 6

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