What is the integral of ((x+2))/(x^4+2x^3+x^2) ?

The integral of ((x+2))/(x^4+2x^3+x^2) is -3ln|x|-2/x+3ln|x+1|-1/(x+1)+C

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integral of ((x+2))/(x^4+2x^3+x^2)

\int \frac{( x + 2 )}{x ^{4} + 2 x ^{3} + x ^{2}} d x

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

Solution steps

\int \frac{x + 2}{x ^{4} + 2 x ^{3} + x ^{2}} d x

Take the partial fraction of

\frac{x + 2}{x ^{4} + 2 x ^{3} + x ^{2}} : - \frac{3}{x} + \frac{2}{x ^{2}} + \frac{3}{x + 1} + \frac{1}{( x + 1 ) ^{2}}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

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

Add a constant to the solution

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

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What is the integral of ((x+2))/(x^4+2x^3+x^2) ?
The integral of ((x+2))/(x^4+2x^3+x^2) is -3ln|x|-2/x+3ln|x+1|-1/(x+1)+C