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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} ((x cos (x)) (cos (x) + tan (x)))

\int_{0}^{1} 1 - x^{\frac{2}{7}} d x

x^{2} - 3 x y + x y^{^{'}} = 0

\frac{d}{d t} (cos (t))

\frac{d}{d x} (ln (\frac{5 x + 1}{5 x - 1}))

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