What is the integral of 1/((t-4)^2) ?

The integral of 1/((t-4)^2) is -1/(t-4)+C

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integral of 1/((t-4)^2)

\int \frac{1}{( t - 4 ) ^{2}} d t

- \frac{1}{t - 4} + C

Solution steps

\int \frac{1}{( t - 4 ) ^{2}} d t

Apply u-substitution

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

Apply the Power Rule

= - \frac{1}{u}

Substitute back

u = t - 4

= - \frac{1}{t - 4}

Add a constant to the solution

= - \frac{1}{t - 4} + C

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