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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{1}{s ( ( s + 2 ) ^{2} + 1 )}

n = 1 \sum \infty 6 e^{- 8 n}

x \to 0 lim (2 - x)

\int - e^{- s t} d t

\int \frac{2}{e ^{t}} d t

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