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

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

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

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

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

Solution steps

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

Complete the square

x^{2} - 4 x + 3 : (x - 2)^{2} - 1

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

Apply u-substitution

: \int \frac{u + 3}{u ^{2} - 1} d u

= \int \frac{u + 3}{u ^{2} - 1} d u

Expand

\frac{u + 3}{u ^{2} - 1} : \frac{u}{u ^{2} - 1} + \frac{3}{u ^{2} - 1}

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}{u ^{2} - 1} d u + \int \frac{3}{u ^{2} - 1} d u

\int \frac{u}{u ^{2} - 1} d u = \frac{1}{2} ln u^{2} - 1

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

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

Substitute back

u = x - 2

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

Simplify

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

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

Add a constant to the solution

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

y = x^{2} - 2, y = 7

f (x) = \frac{- 6 x}{x ^{2} + 1}, (0, 0)

x^{2} + 4

\int \frac{1}{sin ( \frac{y}{2} )} d y

\int_{0}^{t} \frac{1}{1 + r} d r

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