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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

\frac{1}{( x + 2 ) ( x + 1 )} : - \frac{1}{x + 2} + \frac{1}{x + 1}

= \int_{- 1}^{4} - \frac{1}{x + 2} + \frac{1}{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_{- 1}^{4} \frac{1}{x + 2} d x + \int_{- 1}^{4} \frac{1}{x + 1} d x

\int_{- 1}^{4} \frac{1}{x + 2} d x = ln (6)

\int_{- 1}^{4} \frac{1}{x + 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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