What is the integral from 2 to 4 of 1/((x-4)(x-1)) ?

The integral from 2 to 4 of 1/((x-4)(x-1)) is diverges

Popular >

integral from 2 to 4 of 1/((x-4)(x-1))

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

diverges

Solution steps

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

Take the partial fraction of

\frac{1}{( x - 4 ) ( x - 1 )} : \frac{1}{3 ( x - 4 )} - \frac{1}{3 ( x - 1 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{9}^{16} \frac{1}{2 - x} d x

\frac{\partial}{\partial y} (sin (y) + y, x)

\int_{0}^{2} t e^{t} d t

sin^{2} (x), sin^{3} (x), 0, π

\int 3 sin (x) + 2 - 3 e^{x} d x

What is the integral from 2 to 4 of 1/((x-4)(x-1)) ?
The integral from 2 to 4 of 1/((x-4)(x-1)) is diverges