What is the integral from 0 to t of 1/((3-2x)^2) ?

The integral from 0 to t of 1/((3-2x)^2) is t/(3(-2t+3))

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integral from 0 to t of 1/((3-2x)^2)

\int_{0}^{t} \frac{1}{( 3 - 2 x ) ^{2}} d x

\frac{t}{3 ( - 2 t + 3 )}

Solution steps

\int_{0}^{t} \frac{1}{( 3 - 2 x ) ^{2}} d x

Apply u-substitution

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

= \int_{3}^{3 - 2 t} - \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

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

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{3 - 2 t} + \frac{1}{3}

= - \frac{1}{2} (- \frac{1}{3 - 2 t} + \frac{1}{3})

Simplify

= \frac{t}{3 ( - 2 t + 3 )}

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What is the integral from 0 to t of 1/((3-2x)^2) ?
The integral from 0 to t of 1/((3-2x)^2) is t/(3(-2t+3))