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

The integral of (x+1)/(sqrt(x^2-3x+2)) is sqrt(x^2-3x+2)+5/2 ln|2sqrt(x^2-3x+2)+2x-3|+C

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

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

x^{2} - 3 x + 2 + \frac{5}{2} ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 v + 5}{4 v ^{2} - 1} d v

= \int \frac{2 v + 5}{4 v ^{2} - 1} d v

Expand

\frac{2 v + 5}{4 v ^{2} - 1} : \frac{2 v}{4 v ^{2} - 1} + \frac{5}{4 v ^{2} - 1}

= \int \frac{2 v}{4 v ^{2} - 1} + \frac{5}{4 v ^{2} - 1} d v

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{2 v}{4 v ^{2} - 1} d v + \int \frac{5}{4 v ^{2} - 1} d v

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

\int \frac{5}{4 v ^{2} - 1} d v = \frac{5}{2} ln 4 v^{2} - 1 + 2 v

= \frac{1}{2} 4 v^{2} - 1 + \frac{5}{2} ln 4 v^{2} - 1 + 2 v

Substitute back

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

Simplify

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

= x^{2} - 3 x + 2 + \frac{5}{2} ln 2 x^{2} - 3 x + 2 + 2 x - 3

Add a constant to the solution

= x^{2} - 3 x + 2 + \frac{5}{2} ln 2 x^{2} - 3 x + 2 + 2 x - 3 + C

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What is the integral of (x+1)/(sqrt(x^2-3x+2)) ?
The integral of (x+1)/(sqrt(x^2-3x+2)) is sqrt(x^2-3x+2)+5/2 ln|2sqrt(x^2-3x+2)+2x-3|+C