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

The integral of (x^2-x-8)/((2x-3)(x^2+2x+2)) is 1/2 (ln(4|x^2+2x+2|)+2arctan(x+1)-ln|2x-3|)+C

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 4 u - 29}{2 u ( u ^{2} + 10 u + 29 )} d u

= \int \frac{u ^{2} + 4 u - 29}{2 u ( u ^{2} + 10 u + 29 )} 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} + 4 u - 29}{u ( u ^{2} + 10 u + 29 )} d u

Expand

\frac{u ^{2} + 4 u - 29}{u ( u ^{2} + 10 u + 29 )} : \frac{u ^{2}}{u ( u ^{2} + 10 u + 29 )} + \frac{4 u}{u ( u ^{2} + 10 u + 29 )} - \frac{29}{u ( u ^{2} + 10 u + 29 )}

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 ^{2} + 10 u + 29 )} d u + \int \frac{4 u}{u ( u ^{2} + 10 u + 29 )} d u - \int \frac{29}{u ( u ^{2} + 10 u + 29 )} d u)

\int \frac{u ^{2}}{u ( u ^{2} + 10 u + 29 )} d u = \frac{1}{2} ln u^{2} + 10 u + 29 - \frac{5}{2} arctan (\frac{1}{2} (u + 5))

\int \frac{4 u}{u ( u ^{2} + 10 u + 29 )} d u = 2 arctan (\frac{1}{2} (u + 5))

\int \frac{29}{u ( u ^{2} + 10 u + 29 )} d u = ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 10 u + 29 - \frac{5}{2} arctan (\frac{1}{2} (u + 5))

= \frac{1}{2} (\frac{1}{2} ln u^{2} + 10 u + 29 - \frac{5}{2} arctan (\frac{1}{2} (u + 5)) + 2 arctan (\frac{1}{2} (u + 5)) - (ln ∣ u ∣ - \frac{1}{2} ln u^{2} + 10 u + 29 - \frac{5}{2} arctan (\frac{1}{2} (u + 5))))

Substitute back

u = 2 x - 3

= \frac{1}{2} (\frac{1}{2} ln (2 x - 3)^{2} + 10 (2 x - 3) + 29 - \frac{5}{2} arctan (\frac{1}{2} (2 x - 3 + 5)) + 2 arctan (\frac{1}{2} (2 x - 3 + 5)) - (ln ∣ 2 x - 3 ∣ - \frac{1}{2} ln (2 x - 3)^{2} + 10 (2 x - 3) + 29 - \frac{5}{2} arctan (\frac{1}{2} (2 x - 3 + 5))))

Simplify

\frac{1}{2} (\frac{1}{2} ln (2 x - 3)^{2} + 10 (2 x - 3) + 29 - \frac{5}{2} arctan (\frac{1}{2} (2 x - 3 + 5)) + 2 arctan (\frac{1}{2} (2 x - 3 + 5)) - (ln ∣ 2 x - 3 ∣ - \frac{1}{2} ln (2 x - 3)^{2} + 10 (2 x - 3) + 29 - \frac{5}{2} arctan (\frac{1}{2} (2 x - 3 + 5)))) : \frac{1}{2} (ln (4 x^{2} + 2 x + 2) + 2 arctan (x + 1) - ln ∣ 2 x - 3 ∣)

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

Add a constant to the solution

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

\int - \frac{7}{64 + 36 t ^{2}} d t

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

x \to 0 + lim (ln (\frac{2}{x}))

x \to 0 lim ((\frac{1}{x ^{2}})^{2})

\int \frac{x ^{3}}{9 + x ^{8}} d x

What is the integral of (x^2-x-8)/((2x-3)(x^2+2x+2)) ?
The integral of (x^2-x-8)/((2x-3)(x^2+2x+2)) is 1/2 (ln(4|x^2+2x+2|)+2arctan(x+1)-ln|2x-3|)+C