What is the integral from 1 to 3 of (x+2)(x-1)^5 ?

The integral from 1 to 3 of (x+2)(x-1)^5 is 352/7

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integral from 1 to 3 of (x+2)(x-1)^5

\int_{1}^{3} (x + 2) (x - 1)^{5} d x

\frac{352}{7}

Decimal Notation

50.28571 \dots

Solution steps

\int_{1}^{3} (x + 2) (x - 1)^{5} d x

Multiply by the conjugate of

x + 2 : \frac{x ^{2} - 4}{x - 2}

= \int_{1}^{3} \frac{x ^{2} - 4}{x - 2} (x - 1)^{5} d x

Find undefined (singularity) points:

x = 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_{1}^{2} \frac{x ^{2} - 4}{x - 2} (x - 1)^{5} d x + \int_{2}^{3} \frac{x ^{2} - 4}{x - 2} (x - 1)^{5} d x

\int_{1}^{2} \frac{x ^{2} - 4}{x - 2} (x - 1)^{5} d x = \frac{9}{14}

\int_{2}^{3} \frac{x ^{2} - 4}{x - 2} (x - 1)^{5} d x = \frac{695}{14}

= \frac{9}{14} + \frac{695}{14}

Simplify

\frac{9}{14} + \frac{695}{14} : \frac{352}{7}

= \frac{352}{7}

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

\int e^{x^{3} + 1} x^{2} d x

\frac{d}{d x} (ln (e^{3 x} + e^{- 3 x}))

ln (\frac{x + 3}{x - 1})

\frac{d}{d x} (5 sec (4 x))

What is the integral from 1 to 3 of (x+2)(x-1)^5 ?
The integral from 1 to 3 of (x+2)(x-1)^5 is 352/7