What is the integral of (3x+12)/(x^3+4x^2+4x) ?

The integral of (3x+12)/(x^3+4x^2+4x) is 3ln|x|-3ln|x+2|+3/(x+2)+C

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integral of (3x+12)/(x^3+4x^2+4x)

\int \frac{3 x + 12}{x ^{3} + 4 x ^{2} + 4 x} d x

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

Solution steps

\int \frac{3 x + 12}{x ^{3} + 4 x ^{2} + 4 x} d x

Take the partial fraction of

\frac{3 x + 12}{x ^{3} + 4 x ^{2} + 4 x} : \frac{3}{x} - \frac{3}{x + 2} - \frac{3}{( x + 2 ) ^{2}}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

Simplify

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

Add a constant to the solution

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

f (x) = \frac{1}{2 x}

\int_{- 1}^{2} 3 x^{2} + 6 x d x

f (x) = 1 - x^{2}

\int \frac{1}{( 5 - x ^{2} ) ^{\frac{3}{2}}} d x

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

What is the integral of (3x+12)/(x^3+4x^2+4x) ?
The integral of (3x+12)/(x^3+4x^2+4x) is 3ln|x|-3ln|x+2|+3/(x+2)+C