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

The integral of (x-5)/(x^3+2x^2+x) is -5ln|x|+5ln|x+1|-6/(x+1)+C

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

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

- 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + C

Solution steps

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

Take the partial fraction of

\frac{x - 5}{x ^{3} + 2 x ^{2} + x} : - \frac{5}{x} + \frac{5}{x + 1} + \frac{6}{( x + 1 ) ^{2}}

= \int - \frac{5}{x} + \frac{5}{x + 1} + \frac{6}{( 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{5}{x} d x + \int \frac{5}{x + 1} d x + \int \frac{6}{( x + 1 ) ^{2}} d x

\int \frac{5}{x} d x = 5 ln ∣ x ∣

\int \frac{5}{x + 1} d x = 5 ln ∣ x + 1 ∣

\int \frac{6}{( x + 1 ) ^{2}} d x = - \frac{6}{x + 1}

= - 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{6}{x + 1}

Add a constant to the solution

= - 5 ln ∣ x ∣ + 5 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + C

\frac{\partial}{\partial x} (x ln (z^{5} + x^{4}))

\frac{\partial}{\partial x} (7 x y^{2} + 1)

\frac{d}{d x} (4 ln (2 x))

x \to 0 lim ((1 + 5 x)^{\frac{3}{x}})

What is the integral of (x-5)/(x^3+2x^2+x) ?
The integral of (x-5)/(x^3+2x^2+x) is -5ln|x|+5ln|x+1|-6/(x+1)+C