What is the integral of (x^2-4x+7)/((x+1)(x^2-2x+3)) ?

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

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integral of (x^2-4x+7)/((x+1)(x^2-2x+3))

\int \frac{x ^{2} - 4 x + 7}{( x + 1 ) ( x ^{2} - 2 x + 3 )} d x

\frac{1}{2} ln \frac{3}{2} + \frac{1}{2} x^{2} - x - 22 arctan (\frac{1}{2} (x - 1)) + 2 ln ∣ x + 1 ∣ - ln \frac{3}{2} + \frac{1}{2} x^{2} - x + 22 arctan (\frac{1}{2} (x - 1)) + C

Solution steps

\int \frac{x ^{2} - 4 x + 7}{( x + 1 ) ( x ^{2} - 2 x + 3 )} d x

Apply u-substitution

: \int \frac{u ^{2} - 6 u + 12}{u ( u ^{2} - 4 u + 6 )} d u

= \int \frac{u ^{2} - 6 u + 12}{u ( u ^{2} - 4 u + 6 )} d u

Expand

\frac{u ^{2} - 6 u + 12}{u ( u ^{2} - 4 u + 6 )} : \frac{u ^{2}}{u ( u ^{2} - 4 u + 6 )} - \frac{6 u}{u ( u ^{2} - 4 u + 6 )} + \frac{12}{u ( u ^{2} - 4 u + 6 )}

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} - 4 u + 6 )} d u - \int \frac{6 u}{u ( u ^{2} - 4 u + 6 )} d u + \int \frac{12}{u ( u ^{2} - 4 u + 6 )} d u

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

\int \frac{6 u}{u ( u ^{2} - 4 u + 6 )} d u = 32 arctan (\frac{1}{2} (u - 2))

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

= \frac{1}{2} ln \frac{1}{2} (u - 2)^{2} + 1 + 2 arctan (\frac{1}{2} (u - 2)) - 32 arctan (\frac{1}{2} (u - 2)) + 12 (\frac{1}{6} ln ∣ u ∣ + \frac{1}{6} (- \frac{1}{2} ln \frac{1}{2} u^{2} - 2 u + 3 + 2 arctan (\frac{1}{2} (u - 2))))

Substitute back

u = x + 1

= \frac{1}{2} ln \frac{1}{2} (x + 1 - 2)^{2} + 1 + 2 arctan (\frac{1}{2} (x + 1 - 2)) - 32 arctan (\frac{1}{2} (x + 1 - 2)) + 12 (\frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} (- \frac{1}{2} ln \frac{1}{2} (x + 1)^{2} - 2 (x + 1) + 3 + 2 arctan (\frac{1}{2} (x + 1 - 2))))

Simplify

\frac{1}{2} ln \frac{1}{2} (x + 1 - 2)^{2} + 1 + 2 arctan (\frac{1}{2} (x + 1 - 2)) - 32 arctan (\frac{1}{2} (x + 1 - 2)) + 12 (\frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{6} (- \frac{1}{2} ln \frac{1}{2} (x + 1)^{2} - 2 (x + 1) + 3 + 2 arctan (\frac{1}{2} (x + 1 - 2)))) : \frac{1}{2} ln \frac{3}{2} + \frac{1}{2} x^{2} - x - 22 arctan (\frac{1}{2} (x - 1)) + 2 ln ∣ x + 1 ∣ - ln \frac{3}{2} + \frac{1}{2} x^{2} - x + 22 arctan (\frac{1}{2} (x - 1))

= \frac{1}{2} ln \frac{3}{2} + \frac{1}{2} x^{2} - x - 22 arctan (\frac{1}{2} (x - 1)) + 2 ln ∣ x + 1 ∣ - ln \frac{3}{2} + \frac{1}{2} x^{2} - x + 22 arctan (\frac{1}{2} (x - 1))

Add a constant to the solution

= \frac{1}{2} ln \frac{3}{2} + \frac{1}{2} x^{2} - x - 22 arctan (\frac{1}{2} (x - 1)) + 2 ln ∣ x + 1 ∣ - ln \frac{3}{2} + \frac{1}{2} x^{2} - x + 22 arctan (\frac{1}{2} (x - 1)) + C

y = 5 tan (x)

t^{2} \frac{d y}{d t} + t y = 4

\int_{0}^{1} \frac{3}{2} (x^{2} + y^{2}) d x

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

\frac{\partial}{\partial x} (2 x \cdot sin (9 x^{2} y))

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