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

The integral of 1/((x-3)^3) is -1/(2(x-3)^2)+C

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integral of 1/((x-3)^3)

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

- \frac{1}{2 ( x - 3 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

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

Substitute back

u = x - 3

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

Add a constant to the solution

= - \frac{1}{2 ( x - 3 ) ^{2}} + C

\int 3 - x d x

\frac{3}{( x - 7 ) ^{4}}

\frac{d}{d x} (\frac{x ^{4}}{8} + \frac{x ^{2}}{6} - \frac{x}{5} + \frac{1}{7})

\int e^{c s c (2 x)} csc (2 x) cot (2 x) d x

16 y^{^{''}} - 8 y^{^{'}} + y = 0

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