What is the integral of (x-2)/(x^2-5x+4) ?

The integral of (x-2)/(x^2-5x+4) is 2(1/4 ln|4x^2-20x+16|-1/12 (ln|-2/3+2/3 x|-ln|-8/3+2/3 x|))+C

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integral of (x-2)/(x^2-5x+4)

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

2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{1}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) + C

Solution steps

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

Complete the square

x^{2} - 5 x + 4 : (x - \frac{5}{2})^{2} - \frac{9}{4}

= \int \frac{x - 2}{( x - \frac{5}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 9} d u

= \int \frac{2 ( 2 u + 1 )}{4 u ^{2} - 9} 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 + 1}{4 u ^{2} - 9} d u

Expand

\frac{2 u + 1}{4 u ^{2} - 9} : \frac{2 u}{4 u ^{2} - 9} + \frac{1}{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

= 2 (\int \frac{2 u}{4 u ^{2} - 9} d u + \int \frac{1}{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{1}{4 u ^{2} - 9} d u = - \frac{1}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1)

= 2 (\frac{1}{4} ln 4 u^{2} - 9 - \frac{1}{12} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1))

Substitute back

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

= 2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 9 - \frac{1}{12} (ln \frac{2}{3} (x - \frac{5}{2}) + 1 - ln \frac{2}{3} (x - \frac{5}{2}) - 1))

Simplify

2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 9 - \frac{1}{12} (ln \frac{2}{3} (x - \frac{5}{2}) + 1 - ln \frac{2}{3} (x - \frac{5}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{1}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x))

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{1}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 16 - \frac{1}{12} (ln - \frac{2}{3} + \frac{2}{3} x - ln - \frac{8}{3} + \frac{2}{3} x)) + C

\int \frac{x arcsin ( x ^{2} )}{1 - x ^{4}} d x

\int cos (5 x) e^{s i n (5 x)} d x

\int - 5 x + 15 d x

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

1 - x^{2}

What is the integral of (x-2)/(x^2-5x+4) ?
The integral of (x-2)/(x^2-5x+4) is 2(1/4 ln|4x^2-20x+16|-1/12 (ln|-2/3+2/3 x|-ln|-8/3+2/3 x|))+C