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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\int 4 e^{- t} t d t

\int sec (8 x) d x

f (x) = 6 sec (x)

\frac{\partial}{\partial x} (x^{3} + 8 x^{2} y + 12 y^{2})

\frac{d y}{d x} + 8 y = e^{7 x}

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