What is the integral of ((8+3x-x^2))/((2x+3)(x+2)^2) ?

The integral of ((8+3x-x^2))/((2x+3)(x+2)^2) is ((x+2)(5ln|2x+3|-6ln(2|x+2|))-4)/(2(x+2))+C

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integral of ((8+3x-x^2))/((2x+3)(x+2)^2)

\int \frac{( 8 + 3 x - x ^{2} )}{( 2 x + 3 ) ( x + 2 ) ^{2}} d x

\frac{( x + 2 ) ( 5 ln ∣ 2 x + 3 ∣ - 6 ln ( 2 ∣ x + 2 ∣ ) ) - 4}{2 ( x + 2 )} + C

Solution steps

\int \frac{8 + 3 x - x ^{2}}{( 2 x + 3 ) ( x + 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{- u ^{2} + 12 u + 5}{2 u ( u + 1 ) ^{2}} d u

= \int \frac{- u ^{2} + 12 u + 5}{2 u ( u + 1 ) ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{- u ^{2} + 12 u + 5}{u ( u + 1 ) ^{2}} d u

Expand

\frac{- u ^{2} + 12 u + 5}{u ( u + 1 ) ^{2}} : - \frac{u ^{2}}{u ( u + 1 ) ^{2}} + \frac{12 u}{u ( u + 1 ) ^{2}} + \frac{5}{u ( u + 1 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int \frac{u ^{2}}{u ( u + 1 ) ^{2}} d u + \int \frac{12 u}{u ( u + 1 ) ^{2}} d u + \int \frac{5}{u ( u + 1 ) ^{2}} d u)

\int \frac{u ^{2}}{u ( u + 1 ) ^{2}} d u = ln ∣ u + 1 ∣ + \frac{1}{u + 1}

\int \frac{12 u}{u ( u + 1 ) ^{2}} d u = - \frac{12}{u + 1}

\int \frac{5}{u ( u + 1 ) ^{2}} d u = 5 (ln ∣ u ∣ - ln ∣ u + 1 ∣ + \frac{1}{u + 1})

= \frac{1}{2} (- (ln ∣ u + 1 ∣ + \frac{1}{u + 1}) - \frac{12}{u + 1} + 5 (ln ∣ u ∣ - ln ∣ u + 1 ∣ + \frac{1}{u + 1}))

Substitute back

u = 2 x + 3

= \frac{1}{2} (- (ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}) - \frac{12}{2 x + 3 + 1} + 5 (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}))

Simplify

\frac{1}{2} (- (ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}) - \frac{12}{2 x + 3 + 1} + 5 (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1})) : \frac{( x + 2 ) ( 5 ln ∣ 2 x + 3 ∣ - 6 ln ( 2 ∣ x + 2 ∣ ) ) - 4}{2 ( x + 2 )}

= \frac{( x + 2 ) ( 5 ln ∣ 2 x + 3 ∣ - 6 ln ( 2 ∣ x + 2 ∣ ) ) - 4}{2 ( x + 2 )}

Add a constant to the solution

= \frac{( x + 2 ) ( 5 ln ∣ 2 x + 3 ∣ - 6 ln ( 2 ∣ x + 2 ∣ ) ) - 4}{2 ( x + 2 )} + C

\frac{\partial}{\partial x} (3 e^{x y})

- ln (1 - x)

\frac{d}{d x} (ln (x^{2} + 5 x))

\int_{0}^{1} \frac{1}{3 - 2 x} d x

f (x) = x^{2}, at x = 4

What is the integral of ((8+3x-x^2))/((2x+3)(x+2)^2) ?
The integral of ((8+3x-x^2))/((2x+3)(x+2)^2) is ((x+2)(5ln|2x+3|-6ln(2|x+2|))-4)/(2(x+2))+C