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

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

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integral of (x-1)/(x^3+4x^2+4x)

\int \frac{x - 1}{x ^{3} + 4 x ^{2} + 4 x} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 2 ∣ - \frac{3}{2 ( x + 2 )} + C

Solution steps

\int \frac{x - 1}{x ^{3} + 4 x ^{2} + 4 x} d x

Take the partial fraction of

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

= \int - \frac{1}{4 x} + \frac{1}{4 ( x + 2 )} + \frac{3}{2 ( x + 2 ) ^{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}{4 x} d x + \int \frac{1}{4 ( x + 2 )} d x + \int \frac{3}{2 ( x + 2 ) ^{2}} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{4 ( x + 2 )} d x = \frac{1}{4} ln ∣ x + 2 ∣

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

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 2 ∣ - \frac{3}{2 ( x + 2 )}

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 2 ∣ - \frac{3}{2 ( x + 2 )} + C

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