What is the integral of (6x)/((x-5)^2) ?

The integral of (6x)/((x-5)^2) is 6(ln|x-5|-5/(x-5))+C

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integral of (6x)/((x-5)^2)

\int \frac{6 x}{( x - 5 ) ^{2}} d x

6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5}) + C

Solution steps

\int \frac{6 x}{( x - 5 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{( x - 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{u + 5}{u ^{2}} d u

= 6 \cdot \int \frac{u + 5}{u ^{2}} d u

Expand

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

= 6 \cdot \int \frac{1}{u} + \frac{5}{u ^{2}} 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

= 6 (\int \frac{1}{u} d u + \int \frac{5}{u ^{2}} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 6 (ln ∣ u ∣ - \frac{5}{u})

Substitute back

u = x - 5

= 6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5})

Add a constant to the solution

= 6 (ln ∣ x - 5 ∣ - \frac{5}{x - 5}) + C

x \to 2 lim (6 x^{2} + 1)

\int_{- \infty}^{0} e^{9 x} d x

t^{2} (\frac{d y}{d t}) + 4 t y - y^{3} = 0

x^{3} + y^{3} - 6 x y = 0, (\frac{4}{3}, \frac{8}{3})

\int_{2}^{5} 4 x^{3} d x

What is the integral of (6x)/((x-5)^2) ?
The integral of (6x)/((x-5)^2) is 6(ln|x-5|-5/(x-5))+C