What is the integral of (x^2-3x-7)/((2x+3)(x+1)) ?

The integral of (x^2-3x-7)/((2x+3)(x+1)) is 1/4 (2x+ln|2x+3|-12ln|2x+2|+2)+C

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integral of (x^2-3x-7)/((2x+3)(x+1))

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

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

Solution steps

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

Apply u-substitution

: \int \frac{u ^{2} - 12 u - 1}{4 u ( u - 1 )} d u

= \int \frac{u ^{2} - 12 u - 1}{4 u ( u - 1 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{u ^{2} - 12 u - 1}{u ( u - 1 )} d u

Expand

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

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

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

\int \frac{12 u}{u ( u - 1 )} d u = 12 ln ∣ u - 1 ∣

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

= \frac{1}{4} (u - 1 + ln ∣ u - 1 ∣ - 12 ln ∣ u - 1 ∣ - (- ln ∣ u ∣ + ln ∣ u - 1 ∣))

Substitute back

u = 2 x + 3

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{4} (2 x + ln ∣ 2 x + 3 ∣ - 12 ln ∣ 2 x + 2 ∣ + 2) + C

\int_{1}^{e^{3}} \frac{1}{x ( 1 + ln ( x ) )} d x

y^{^{''}} + 25 y = 4 csc^{2} (5 t)

\frac{\partial}{\partial x} (3 x cos (y))

x \to 0 lim (\frac{1}{cos ( 8 x )})

\frac{d}{d x} (- 11 (4 x^{3} - 11)^{13})

What is the integral of (x^2-3x-7)/((2x+3)(x+1)) ?
The integral of (x^2-3x-7)/((2x+3)(x+1)) is 1/4 (2x+ln|2x+3|-12ln|2x+2|+2)+C