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

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

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

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

- 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{3}{x} + C

Solution steps

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

Expand

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

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

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

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

Simplify

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

= - 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{3}{x}

Add a constant to the solution

= - 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{2}{x + 1} - \frac{3}{x} + C

\frac{\partial}{\partial x} (- \frac{a}{x})

y = x^{2} - 8 x, (3, - 15)

\int \frac{1}{x ^{3} + 8 x} d x

f (x) = 4 cos (2 x)

n = 0 \sum \infty \frac{4}{n}

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