What is the integral from-2 to 3 of (8x+6)/(4x^2+6x) ?

The integral from-2 to 3 of (8x+6)/(4x^2+6x) is diverges

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integral from-2 to 3 of (8x+6)/(4x^2+6x)

\int_{- 2}^{3} \frac{8 x + 6}{4 x ^{2} + 6 x} d x

diverges

Solution steps

\int_{- 2}^{3} \frac{8 x + 6}{4 x ^{2} + 6 x} d x

Find undefined (singularity) points:

x = 0, x = - \frac{3}{2}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{- 2}^{- \frac{3}{2}} \frac{8 x + 6}{4 x ^{2} + 6 x} d x + \int_{- \frac{3}{2}}^{0} \frac{8 x + 6}{4 x ^{2} + 6 x} d x + \int_{0}^{3} \frac{8 x + 6}{4 x ^{2} + 6 x} d x

\int_{- 2}^{- \frac{3}{2}} \frac{8 x + 6}{4 x ^{2} + 6 x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int_{3}^{\infty} e^{- s t} (2 t - 3) d t

What is the integral from-2 to 3 of (8x+6)/(4x^2+6x) ?
The integral from-2 to 3 of (8x+6)/(4x^2+6x) is diverges