What is the integral of (x+1)/(x^2-5x+6) ?

The integral of (x+1)/(x^2-5x+6) is 2(1/4 ln|4x^2-20x+24|-7/4 (ln|2x-4|-ln|2x-6|))+C

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integral of (x+1)/(x^2-5x+6)

\int \frac{x + 1}{x ^{2} - 5 x + 6} d x

2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{7}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

Solution steps

\int \frac{x + 1}{x ^{2} - 5 x + 6} d x

Complete the square

x^{2} - 5 x + 6 : (x - \frac{5}{2})^{2} - \frac{1}{4}

= \int \frac{x + 1}{( x - \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 7 )}{4 u ^{2} - 1} d u

= \int \frac{2 ( 2 u + 7 )}{4 u ^{2} - 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{2 u + 7}{4 u ^{2} - 1} d u

Expand

\frac{2 u + 7}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{7}{4 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

= 2 (\int \frac{2 u}{4 u ^{2} - 1} d u + \int \frac{7}{4 u ^{2} - 1} d u)

\int \frac{2 u}{4 u ^{2} - 1} d u = \frac{1}{4} ln 4 u^{2} - 1

\int \frac{7}{4 u ^{2} - 1} d u = - \frac{7}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - \frac{7}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

Substitute back

u = x - \frac{5}{2}

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

Simplify

2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 1 - \frac{7}{4} (ln 2 (x - \frac{5}{2}) + 1 - ln 2 (x - \frac{5}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{7}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{7}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{7}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

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

\int cos (2 x) sin (5 x) d x

What is the integral of (x+1)/(x^2-5x+6) ?
The integral of (x+1)/(x^2-5x+6) is 2(1/4 ln|4x^2-20x+24|-7/4 (ln|2x-4|-ln|2x-6|))+C