What is the integral of (x^2+7x-6)/((x+1)(x^2-4x+7)) ?

The integral of (x^2+7x-6)/((x+1)(x^2-4x+7)) is

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integral of (x^2+7x-6)/((x+1)(x^2-4x+7))

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 7 )} d x

\frac{1}{2} ln \frac{1}{3} (x - 2)^{2} + 1 + \frac{5}{3} arctan (\frac{1}{3} (x - 2)) - ln ∣ x + 1 ∣ + \frac{1}{2} ln \frac{x ^{2} - 4 x}{3} + \frac{7}{3} + C

Solution steps

\int \frac{x ^{2} + 7 x - 6}{( x + 1 ) ( x ^{2} - 4 x + 7 )} d x

Apply u-substitution

: \int \frac{u ^{2} + 5 u - 12}{u ( u ^{2} - 6 u + 12 )} d u

= \int \frac{u ^{2} + 5 u - 12}{u ( u ^{2} - 6 u + 12 )} d u

Expand

\frac{u ^{2} + 5 u - 12}{u ( u ^{2} - 6 u + 12 )} : \frac{u ^{2}}{u ( u ^{2} - 6 u + 12 )} + \frac{5 u}{u ( u ^{2} - 6 u + 12 )} - \frac{12}{u ( u ^{2} - 6 u + 12 )}

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{u ^{2}}{u ( u ^{2} - 6 u + 12 )} d u + \int \frac{5 u}{u ( u ^{2} - 6 u + 12 )} d u - \int \frac{12}{u ( u ^{2} - 6 u + 12 )} d u

\int \frac{u ^{2}}{u ( u ^{2} - 6 u + 12 )} d u = \frac{1}{2} ln \frac{1}{3} (u - 3)^{2} + 1 + 3 arctan (\frac{1}{3} (u - 3))

\int \frac{5 u}{u ( u ^{2} - 6 u + 12 )} d u = \frac{5}{3} arctan (\frac{1}{3} (u - 3))

\int \frac{12}{u ( u ^{2} - 6 u + 12 )} d u = ln ∣ u ∣ - \frac{1}{2} ln \frac{1}{3} u^{2} - 2 u + 4 + 3 arctan (\frac{1}{3} (u - 3))

= \frac{1}{2} ln \frac{1}{3} (u - 3)^{2} + 1 + 3 arctan (\frac{1}{3} (u - 3)) + \frac{5}{3} arctan (\frac{1}{3} (u - 3)) - (ln ∣ u ∣ - \frac{1}{2} ln \frac{1}{3} u^{2} - 2 u + 4 + 3 arctan (\frac{1}{3} (u - 3)))

Substitute back

u = x + 1

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

Simplify

\frac{1}{2} ln \frac{1}{3} (x + 1 - 3)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1 - 3)) + \frac{5}{3} arctan (\frac{1}{3} (x + 1 - 3)) - (ln ∣ x + 1 ∣ - \frac{1}{2} ln \frac{1}{3} (x + 1)^{2} - 2 (x + 1) + 4 + 3 arctan (\frac{1}{3} (x + 1 - 3))) : \frac{1}{2} ln \frac{1}{3} (x - 2)^{2} + 1 + \frac{5}{3} arctan (\frac{1}{3} (x - 2)) - ln ∣ x + 1 ∣ + \frac{1}{2} ln \frac{x ^{2} - 4 x}{3} + \frac{7}{3}

= \frac{1}{2} ln \frac{1}{3} (x - 2)^{2} + 1 + \frac{5}{3} arctan (\frac{1}{3} (x - 2)) - ln ∣ x + 1 ∣ + \frac{1}{2} ln \frac{x ^{2} - 4 x}{3} + \frac{7}{3}

Add a constant to the solution

= \frac{1}{2} ln \frac{1}{3} (x - 2)^{2} + 1 + \frac{5}{3} arctan (\frac{1}{3} (x - 2)) - ln ∣ x + 1 ∣ + \frac{1}{2} ln \frac{x ^{2} - 4 x}{3} + \frac{7}{3} + C

\int \frac{8}{θ ^{2}} cos (\frac{8}{θ}) d θ

y = ln (8 - 4 x)

f (x) = 21 x

y = 8 x sin (x), (\frac{π}{2}, 4 π)

y = \frac{1}{x ^{5}}, (1, 1)

What is the integral of (x^2+7x-6)/((x+1)(x^2-4x+7)) ?
The integral of (x^2+7x-6)/((x+1)(x^2-4x+7)) is