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

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

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

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

\frac{1}{2}

Decimal Notation

0.5

Solution steps

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

Apply u-substitution

: \int_{- 2}^{- 1} \frac{1}{u ^{2}} d u

= \int_{- 2}^{- 1} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int_{1}^{y} \frac{1}{y ^{2}} d y

\int_{0}^{2} \frac{1}{1 + x ^{3}} d x

\int_{0}^{\frac{π}{2}} ln (x) d x

\int_{0}^{6} (x^{2} + x + 1) d x

\int_{1}^{5} x^{3} + 8 x - 4 d x

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