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

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

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integral from k to 2k of 2/((2x-k)^2)

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

\frac{2}{3 k}

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Simplify

2 \cdot \frac{1}{2} [- \frac{1}{u}]_{k}^{3 k} : [- \frac{1}{u}]_{k}^{3 k}

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

Compute the boundaries:

- \frac{1}{3 k} + \frac{1}{k}

= - \frac{1}{3 k} + \frac{1}{k}

Simplify

= \frac{2}{3 k}

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