What is the integral of (2x-1)/(x^2-7x+10) ?

The integral of (2x-1)/(x^2-7x+10) is ln|4x^2-28x+40|-2ln| 2/3 x-4/3 |+2ln| 2/3 x-10/3 |+C

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integral of (2x-1)/(x^2-7x+10)

\int \frac{2 x - 1}{x ^{2} - 7 x + 10} d x

ln 4 x^{2} - 28 x + 40 - 2 ln \frac{2}{3} x - \frac{4}{3} + 2 ln \frac{2}{3} x - \frac{10}{3} + C

Solution steps

\int \frac{2 x - 1}{x ^{2} - 7 x + 10} d x

Complete the square

x^{2} - 7 x + 10 : (x - \frac{7}{2})^{2} - \frac{9}{4}

= \int \frac{2 x - 1}{( x - \frac{7}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

= \int \frac{4 ( 2 u + 6 )}{4 u ^{2} - 9} d u

Take the constant out:

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

= 4 \cdot \int \frac{2 u + 6}{4 u ^{2} - 9} d u

Expand

\frac{2 u + 6}{4 u ^{2} - 9} : \frac{2 u}{4 u ^{2} - 9} + \frac{6}{4 u ^{2} - 9}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int \frac{2 u}{4 u ^{2} - 9} d u + \int \frac{6}{4 u ^{2} - 9} d u)

\int \frac{2 u}{4 u ^{2} - 9} d u = \frac{1}{4} ln 4 u^{2} - 9

\int \frac{6}{4 u ^{2} - 9} d u = - \frac{1}{2} ln \frac{2}{3} u + 1 + \frac{1}{2} ln \frac{2}{3} u - 1

= 4 (\frac{1}{4} ln 4 u^{2} - 9 - \frac{1}{2} ln \frac{2}{3} u + 1 + \frac{1}{2} ln \frac{2}{3} u - 1)

Substitute back

u = x - \frac{7}{2}

= 4 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 9 - \frac{1}{2} ln \frac{2}{3} (x - \frac{7}{2}) + 1 + \frac{1}{2} ln \frac{2}{3} (x - \frac{7}{2}) - 1)

Simplify

4 (\frac{1}{4} ln 4 (x - \frac{7}{2})^{2} - 9 - \frac{1}{2} ln \frac{2}{3} (x - \frac{7}{2}) + 1 + \frac{1}{2} ln \frac{2}{3} (x - \frac{7}{2}) - 1) : ln 4 x^{2} - 28 x + 40 - 2 ln \frac{2}{3} x - \frac{4}{3} + 2 ln \frac{2}{3} x - \frac{10}{3}

= ln 4 x^{2} - 28 x + 40 - 2 ln \frac{2}{3} x - \frac{4}{3} + 2 ln \frac{2}{3} x - \frac{10}{3}

Add a constant to the solution

= ln 4 x^{2} - 28 x + 40 - 2 ln \frac{2}{3} x - \frac{4}{3} + 2 ln \frac{2}{3} x - \frac{10}{3} + C

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What is the integral of (2x-1)/(x^2-7x+10) ?
The integral of (2x-1)/(x^2-7x+10) is ln|4x^2-28x+40|-2ln| 2/3 x-4/3 |+2ln| 2/3 x-10/3 |+C