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

The integral from 2 to 4 of (x-2)/(x^2-5x+4) is diverges

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

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

diverges

Solution steps

\int_{2}^{4} \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_{2}^{4} \frac{x - 2}{( x - \frac{5}{2} ) ^{2} - \frac{9}{4}} d x

Apply u-substitution

: \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{2 ( 2 u + 1 )}{4 u ^{2} - 9} d u

= \int_{- \frac{1}{2}}^{\frac{3}{2}} \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{1}{2}}^{\frac{3}{2}} \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{1}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 9} d u + \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} - 9} d u)

\int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{2 u}{4 u ^{2} - 9} d u =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{1}^{\infty} e^{- s t} t d t

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