What is the integral of 1/((2-5t)^2) ?

The integral of 1/((2-5t)^2) is 1/(5(2-5t))+C

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integral of 1/((2-5t)^2)

\int \frac{1}{( 2 - 5 t ) ^{2}} d t

\frac{1}{5 ( 2 - 5 t )} + C

Solution steps

\int \frac{1}{( 2 - 5 t ) ^{2}} d t

Apply u-substitution

: \int - \frac{1}{5 u ^{2}} d u

= \int - \frac{1}{5 u ^{2}} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{5} (- \frac{1}{u})

Substitute back

u = 2 - 5 t

= - \frac{1}{5} (- \frac{1}{2 - 5 t})

Simplify

- \frac{1}{5} (- \frac{1}{2 - 5 t}) : \frac{1}{5 ( 2 - 5 t )}

= \frac{1}{5 ( 2 - 5 t )}

Add a constant to the solution

= \frac{1}{5 ( 2 - 5 t )} + C

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What is the integral of 1/((2-5t)^2) ?
The integral of 1/((2-5t)^2) is 1/(5(2-5t))+C