What is the integral of (12x)/((x-1)(x^2+2x+3)) ?

The integral of (12x)/((x-1)(x^2+2x+3)) is 2ln|x-1|-ln|x^2+2x+3|+4sqrt(2)arctan(1/(sqrt(2))(x+1))+C

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integral of (12x)/((x-1)(x^2+2x+3))

\int \frac{12 x}{( x - 1 ) ( x ^{2} + 2 x + 3 )} d x

2 ln ∣ x - 1 ∣ - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)) + C

Solution steps

\int \frac{12 x}{( x - 1 ) ( x ^{2} + 2 x + 3 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int \frac{x}{( x - 1 ) ( x ^{2} + 2 x + 3 )} d x

Take the partial fraction of

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

= 12 \cdot \int \frac{1}{6 ( x - 1 )} + \frac{- x + 3}{6 ( x ^{2} + 2 x + 3 )} 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

= 12 (\int \frac{1}{6 ( x - 1 )} d x + \int \frac{- x + 3}{6 ( x ^{2} + 2 x + 3 )} d x)

\int \frac{1}{6 ( x - 1 )} d x = \frac{1}{6} ln ∣ x - 1 ∣

\int \frac{- x + 3}{6 ( x ^{2} + 2 x + 3 )} d x = \frac{1}{12} (- ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)))

= 12 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{12} (- ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1))))

Simplify

12 (\frac{1}{6} ln ∣ x - 1 ∣ + \frac{1}{12} (- ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)))) : 2 ln ∣ x - 1 ∣ - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1))

= 2 ln ∣ x - 1 ∣ - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1))

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - ln x^{2} + 2 x + 3 + 42 arctan (\frac{1}{2} (x + 1)) + C

(1.2) (3.3)

\frac{d P}{d t} = 4 Pt

\frac{d}{d x} (e^{s i n (x)} \cdot cos (x))

\int x y x^{2} + y^{2} d y

What is the integral of (12x)/((x-1)(x^2+2x+3)) ?
The integral of (12x)/((x-1)(x^2+2x+3)) is 2ln|x-1|-ln|x^2+2x+3|+4sqrt(2)arctan(1/(sqrt(2))(x+1))+C