What is the integral of 2/((x+1)(x+2)) ?

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

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integral of 2/((x+1)(x+2))

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

2 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

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

= 2 \cdot \int \frac{1}{x + 1} - \frac{1}{x + 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

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

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

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

= 2 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣)

Add a constant to the solution

= 2 (ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣) + C

n = 1 \sum \infty 2 (0.4)^{n - 1}

\int x (x^{2} - 36)^{\frac{1}{2}} d x

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

\frac{\partial}{\partial x} (- 3 x^{2} y^{2})

y = x^{2} + 4, (- 2, 8)

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