What is the integral from 1 to 2 of 1/((1-6x)^4) ?

The integral from 1 to 2 of 1/((1-6x)^4) is 67/166375

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integral from 1 to 2 of 1/((1-6x)^4)

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

\frac{67}{166375}

Decimal

0.00040 \dots

Solution steps

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

Apply u-substitution

: \int_{- 5}^{- 11} - \frac{1}{6 u ^{4}} d u

= \int_{- 5}^{- 11} - \frac{1}{6 u ^{4}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 11}^{- 5} - \frac{1}{6 u ^{4}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{6} \cdot \int_{- 11}^{- 5} \frac{1}{u ^{4}} d u)

Apply the Power Rule

: [- \frac{1}{3 u ^{3}}]_{- 11}^{- 5}

= - (- \frac{1}{6} [- \frac{1}{3 u ^{3}}]_{- 11}^{- 5})

Simplify

= \frac{1}{6} [- \frac{1}{3 u ^{3}}]_{- 11}^{- 5}

Compute the boundaries:

\frac{402}{166375}

= \frac{1}{6} \cdot \frac{402}{166375}

Simplify

= \frac{67}{166375}

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What is the integral from 1 to 2 of 1/((1-6x)^4) ?
The integral from 1 to 2 of 1/((1-6x)^4) is 67/166375