What is the integral of (8sqrt(x))/(x-1) ?

The integral of (8sqrt(x))/(x-1) is -16(1/2 ln|sqrt(x)+1|-1/2 ln|sqrt(x)-1|-sqrt(x))+C

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integral of (8sqrt(x))/(x-1)

\int \frac{8 x}{x - 1} d x

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

Solution steps

\int \frac{8 x}{x - 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{x - 1} d x

Apply u-substitution

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

= 8 \cdot \int \frac{2 u ^{2}}{u ^{2} - 1} d u

Take the constant out:

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

= 8 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{2} - 1} d u

Factor

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

= 8 \cdot 2 \cdot \int \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot 2 (- \int \frac{u ^{2}}{- u ^{2} + 1} d u)

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

= 8 \cdot 2 (- \int \frac{1}{- u ^{2} + 1} - 1 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

= 8 \cdot 2 (- (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

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

\int 1 d u = u

= 8 \cdot 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2} - u))

Substitute back

u = x

= 8 \cdot 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x))

Simplify

8 \cdot 2 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x)) : - 16 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x)

= - 16 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x)

Add a constant to the solution

= - 16 (\frac{1}{2} ln x + 1 - \frac{1}{2} ln x - 1 - x) + C

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What is the integral of (8sqrt(x))/(x-1) ?
The integral of (8sqrt(x))/(x-1) is -16(1/2 ln|sqrt(x)+1|-1/2 ln|sqrt(x)-1|-sqrt(x))+C