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

The integral from 1 to 2 of 1/((3x-1)^2) is 1/10

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

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

\frac{1}{10}

Decimal

0.1

Solution steps

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

Apply u-substitution

: \int_{2}^{5} \frac{1}{3 u ^{2}} d u

= \int_{2}^{5} \frac{1}{3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{2}^{5} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2}^{5}

= \frac{1}{3} [- \frac{1}{u}]_{2}^{5}

Compute the boundaries:

\frac{3}{10}

= \frac{1}{3} \cdot \frac{3}{10}

Simplify

= \frac{1}{10}

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What is the integral from 1 to 2 of 1/((3x-1)^2) ?
The integral from 1 to 2 of 1/((3x-1)^2) is 1/10