What is the integral of (4x)/((x-5)^3) ?

The integral of (4x)/((x-5)^3) is 4(-1/(x-5)-5/(2(x-5)^2))+C

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integral of (4x)/((x-5)^3)

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

4 (- \frac{1}{x - 5} - \frac{5}{2 ( x - 5 ) ^{2}}) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{x}{( x - 5 ) ^{3}} d x

Apply u-substitution

: \int \frac{u + 5}{u ^{3}} d u

= 4 \cdot \int \frac{u + 5}{u ^{3}} d u

Expand

\frac{u + 5}{u ^{3}} : \frac{1}{u ^{2}} + \frac{5}{u ^{3}}

= 4 \cdot \int \frac{1}{u ^{2}} + \frac{5}{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

= 4 (\int \frac{1}{u ^{2}} d u + \int \frac{5}{u ^{3}} d u)

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

\int \frac{5}{u ^{3}} d u = - \frac{5}{2 u ^{2}}

= 4 (- \frac{1}{u} - \frac{5}{2 u ^{2}})

Substitute back

u = x - 5

= 4 (- \frac{1}{x - 5} - \frac{5}{2 ( x - 5 ) ^{2}})

Add a constant to the solution

= 4 (- \frac{1}{x - 5} - \frac{5}{2 ( x - 5 ) ^{2}}) + C

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What is the integral of (4x)/((x-5)^3) ?
The integral of (4x)/((x-5)^3) is 4(-1/(x-5)-5/(2(x-5)^2))+C