What is the integral of x/((x-4)^3) ?

The integral of x/((x-4)^3) is -1/(x-4)-2/((x-4)^2)+C

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integral of x/((x-4)^3)

\int \frac{x}{( x - 4 ) ^{3}} d x

- \frac{1}{x - 4} - \frac{2}{( x - 4 ) ^{2}} + C

Solution steps

\int \frac{x}{( x - 4 ) ^{3}} d x

Apply u-substitution

: \int \frac{u + 4}{u ^{3}} d u

= \int \frac{u + 4}{u ^{3}} d u

Expand

\frac{u + 4}{u ^{3}} : \frac{1}{u ^{2}} + \frac{4}{u ^{3}}

= \int \frac{1}{u ^{2}} + \frac{4}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{2}} d u + \int \frac{4}{u ^{3}} d u

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

\int \frac{4}{u ^{3}} d u = - \frac{2}{u ^{2}}

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

Substitute back

u = x - 4

= - \frac{1}{x - 4} - \frac{2}{( x - 4 ) ^{2}}

Add a constant to the solution

= - \frac{1}{x - 4} - \frac{2}{( x - 4 ) ^{2}} + C

2^{c o s (x)}

\frac{d}{d x} ((5 x + 3)^{4})

\frac{d}{d x} ((arctan (7 x))^{2})

f (x) = 2 x + \frac{10}{x ^{7}}

What is the integral of x/((x-4)^3) ?
The integral of x/((x-4)^3) is -1/(x-4)-2/((x-4)^2)+C