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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

\int \frac{6 x - 5}{x ^{2} - 10 x + 9} d x

\int \frac{x ^{2} + 11 x - 30}{x ^{3} - 5 x ^{2} + 6 x} d x

\int \frac{x}{( 4 x ^{2} + 1 ) ^{5}} d x

\int 5 x x^{2} + 3 d x

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