What is the integral of (3x^2-2x+5)/((x+3)^3) ?

The integral of (3x^2-2x+5)/((x+3)^3) is 3ln|x+3|+(20)/(x+3)-(19)/((x+3)^2)+C

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integral of (3x^2-2x+5)/((x+3)^3)

\int \frac{3 x ^{2} - 2 x + 5}{( x + 3 ) ^{3}} d x

3 ln ∣ x + 3 ∣ + \frac{20}{x + 3} - \frac{19}{( x + 3 ) ^{2}} + C

Solution steps

\int \frac{3 x ^{2} - 2 x + 5}{( x + 3 ) ^{3}} d x

Apply u-substitution

: \int \frac{3 u ^{2} - 20 u + 38}{u ^{3}} d u

= \int \frac{3 u ^{2} - 20 u + 38}{u ^{3}} d u

Expand

\frac{3 u ^{2} - 20 u + 38}{u ^{3}} : \frac{3}{u} - \frac{20}{u ^{2}} + \frac{38}{u ^{3}}

= \int \frac{3}{u} - \frac{20}{u ^{2}} + \frac{38}{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{3}{u} d u - \int \frac{20}{u ^{2}} d u + \int \frac{38}{u ^{3}} d u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int \frac{20}{u ^{2}} d u = - \frac{20}{u}

\int \frac{38}{u ^{3}} d u = - \frac{19}{u ^{2}}

= 3 ln ∣ u ∣ - (- \frac{20}{u}) - \frac{19}{u ^{2}}

Substitute back

u = x + 3

= 3 ln ∣ x + 3 ∣ - (- \frac{20}{x + 3}) - \frac{19}{( x + 3 ) ^{2}}

Simplify

= 3 ln ∣ x + 3 ∣ + \frac{20}{x + 3} - \frac{19}{( x + 3 ) ^{2}}

Add a constant to the solution

= 3 ln ∣ x + 3 ∣ + \frac{20}{x + 3} - \frac{19}{( x + 3 ) ^{2}} + C

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What is the integral of (3x^2-2x+5)/((x+3)^3) ?
The integral of (3x^2-2x+5)/((x+3)^3) is 3ln|x+3|+(20)/(x+3)-(19)/((x+3)^2)+C