What is the integral of (14)/((x-1)^2(x-2)^2) ?

The integral of (14)/((x-1)^2(x-2)^2) is 14(2ln|x-1|-1/(x-1)-2ln|x-2|-1/(x-2))+C

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integral of (14)/((x-1)^2(x-2)^2)

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

14 (2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2}) + C

Solution steps

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

Take the constant out:

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

= 14 \cdot \int \frac{1}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2} ( u - 1 ) ^{2}} d u

= 14 \cdot \int \frac{1}{u ^{2} ( u - 1 ) ^{2}} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u - 1 ) ^{2}} : \frac{2}{u} + \frac{1}{u ^{2}} - \frac{2}{u - 1} + \frac{1}{( u - 1 ) ^{2}}

= 14 \cdot \int \frac{2}{u} + \frac{1}{u ^{2}} - \frac{2}{u - 1} + \frac{1}{( u - 1 ) ^{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

= 14 (\int \frac{2}{u} d u + \int \frac{1}{u ^{2}} d u - \int \frac{2}{u - 1} d u + \int \frac{1}{( u - 1 ) ^{2}} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

\int \frac{2}{u - 1} d u = 2 ln ∣ u - 1 ∣

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

= 14 (2 ln ∣ u ∣ - \frac{1}{u} - 2 ln ∣ u - 1 ∣ - \frac{1}{u - 1})

Substitute back

u = x - 1

= 14 (2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 2 ln ∣ x - 1 - 1 ∣ - \frac{1}{x - 1 - 1})

Simplify

= 14 (2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2})

Add a constant to the solution

= 14 (2 ln ∣ x - 1 ∣ - \frac{1}{x - 1} - 2 ln ∣ x - 2 ∣ - \frac{1}{x - 2}) + C

f (x) = arcsin (4 x)

x \to \infty lim (arctan (x^{2} - x^{5}))

y = e^{x} ln (x), (1, 0)

\frac{\partial}{\partial y} (e^{x} y^{2})

\frac{d}{d x} (tan (x) - 1)

What is the integral of (14)/((x-1)^2(x-2)^2) ?
The integral of (14)/((x-1)^2(x-2)^2) is 14(2ln|x-1|-1/(x-1)-2ln|x-2|-1/(x-2))+C