What is the integral from-2 to 9 of 7/((4x-3)^2) ?

The integral from-2 to 9 of 7/((4x-3)^2) is diverges

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integral from-2 to 9 of 7/((4x-3)^2)

\int_{- 2}^{9} \frac{7}{( 4 x - 3 ) ^{2}} d x

diverges

Solution steps

\int_{- 2}^{9} \frac{7}{( 4 x - 3 ) ^{2}} d x

Find undefined (singularity) points:

x = \frac{3}{4}

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}{4}} \frac{7}{( 4 x - 3 ) ^{2}} d x + \int_{\frac{3}{4}}^{9} \frac{7}{( 4 x - 3 ) ^{2}} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int 24 y (x - y x - \frac{x ^{2}}{2}) d y

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2 x e^{x}

\int 2 t e^{- t^{2}} d t

\int_{0}^{1} x 1 - x^{2} d x

What is the integral from-2 to 9 of 7/((4x-3)^2) ?
The integral from-2 to 9 of 7/((4x-3)^2) is diverges