What is the integral of 1/((z+4)^4) ?

The integral of 1/((z+4)^4) is -1/(3(z+4)^3)+C

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integral of 1/((z+4)^4)

\int \frac{1}{( z + 4 ) ^{4}} d z

- \frac{1}{3 ( z + 4 ) ^{3}} + C

Solution steps

\int \frac{1}{( z + 4 ) ^{4}} d z

Apply u-substitution

: \int \frac{1}{u ^{4}} d u

= \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= - \frac{1}{3 u ^{3}}

Substitute back

u = z + 4

= - \frac{1}{3 ( z + 4 ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( z + 4 ) ^{3}} + C

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What is the integral of 1/((z+4)^4) ?
The integral of 1/((z+4)^4) is -1/(3(z+4)^3)+C