What is the integral from 0 to 1 of (9x+2)^{30} ?

The integral from 0 to 1 of (9x+2)^{30} is (-2^{31}+11^{31})/(279)

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integral from 0 to 1 of (9x+2)^{30}

\int_{0}^{1} (9 x + 2)^{30} d x

\frac{- 2 ^{31} + 1 1 ^{31}}{279}

Decimal Notation

6.87969 E 29

Solution steps

\int_{0}^{1} (9 x + 2)^{30} d x

Multiply by the conjugate of

9 x + 2 : \frac{81 x ^{2} - 4}{9 x - 2}

= \int_{0}^{1} (\frac{81 x ^{2} - 4}{9 x - 2})^{30} d x

Find undefined (singularity) points:

x = \frac{2}{9}

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_{0}^{\frac{2}{9}} (\frac{81 x ^{2} - 4}{9 x - 2})^{30} d x + \int_{\frac{2}{9}}^{1} (\frac{81 x ^{2} - 4}{9 x - 2})^{30} d x

\int_{0}^{\frac{2}{9}} (\frac{81 x ^{2} - 4}{9 x - 2})^{30} d x = \frac{4 ^{31} - 2 ^{31}}{279}

\int_{\frac{2}{9}}^{1} (\frac{81 x ^{2} - 4}{9 x - 2})^{30} d x = \frac{1 1 ^{31} - 4 ^{31}}{279}

= \frac{4 ^{31} - 2 ^{31}}{279} + \frac{1 1 ^{31} - 4 ^{31}}{279}

Simplify

\frac{4 ^{31} - 2 ^{31}}{279} + \frac{1 1 ^{31} - 4 ^{31}}{279} : \frac{- 2 ^{31} + 1 1 ^{31}}{279}

= \frac{- 2 ^{31} + 1 1 ^{31}}{279}

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What is the integral from 0 to 1 of (9x+2)^{30} ?
The integral from 0 to 1 of (9x+2)^{30} is (-2^{31}+11^{31})/(279)