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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

f (x) = (2 x - 2)^{4} (x^{2} + x + 1)^{5}

\frac{\partial}{\partial x} (5 y e^{x})

y = 6 x, (1, 6)

\int sin (x + 7) d x

(- 1.1) (- 4.1)

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