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

The integral from 1 to 2 of 4/((x-1)^{1/4)} is 16/3

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

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

\frac{16}{3}

Decimal

5.33333 \dots

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{( x - 1 ) ^{\frac{1}{4}}} = (x - 1)^{- \frac{1}{4}}

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

Apply u-substitution

= 4 \cdot \int_{0}^{1} \frac{1}{u ^{\frac{1}{4}}} d u

Apply the Power Rule

= 4 [\frac{4}{3} u^{\frac{3}{4}}]_{0}^{1}

Compute the boundaries:

\frac{4}{3}

= 4 \cdot \frac{4}{3}

Simplify

= \frac{16}{3}

What is the integral from 1 to 2 of 4/((x-1)^{1/4)} ?
The integral from 1 to 2 of 4/((x-1)^{1/4)} is 16/3