What is the integral from 0 to 3 of x/(x^2-7x+12) ?

The integral from 0 to 3 of x/(x^2-7x+12) is diverges

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integral from 0 to 3 of x/(x^2-7x+12)

\int_{0}^{3} \frac{x}{x ^{2} - 7 x + 12} d x

diverges

Solution steps

\int_{0}^{3} \frac{x}{x ^{2} - 7 x + 12} d x

Complete the square

x^{2} - 7 x + 12 : (x - \frac{7}{2})^{2} - \frac{1}{4}

= \int_{0}^{3} \frac{x}{( x - \frac{7}{2} ) ^{2} - \frac{1}{4}} d x

Apply u-substitution

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

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

\int_{- \frac{7}{2}}^{- \frac{1}{2}} \frac{2 u}{4 u ^{2} - 1} d u =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 3 of x/(x^2-7x+12) ?
The integral from 0 to 3 of x/(x^2-7x+12) is diverges