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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2} - u + 2}{u ^{3}} d u

= \int \frac{2 u ^{2} - u + 2}{u ^{3}} d u

Expand

\frac{2 u ^{2} - u + 2}{u ^{3}} : \frac{2}{u} - \frac{1}{u ^{2}} + \frac{2}{u ^{3}}

= \int \frac{2}{u} - \frac{1}{u ^{2}} + \frac{2}{u ^{3}} d u

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}{u} d u - \int \frac{1}{u ^{2}} d u + \int \frac{2}{u ^{3}} d u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{2}{u ^{3}} d u = - \frac{1}{u ^{2}}

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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