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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{x - 1}{x ^{3} + x ^{2}} : \frac{2}{x} - \frac{1}{x ^{2}} - \frac{2}{x + 1}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{2}{x + 1} d x = 2 ln ∣ x + 1 ∣

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

Simplify

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

Add a constant to the solution

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

(1 - x^{2}) y^{^{'}} + x y = x

\int \frac{2 x}{k ^{2}} d x

x \to \infty lim (x^{2} - 4 x)

f (x) = 4 sin^{3} (x)

y^{^{'}} + \frac{2}{t} y = \frac{sin ( t )}{t}

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