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

The integral of (8x)/(sqrt(x-1)) is 16(1/3 (x-1)^{3/2}+sqrt(x-1))+C

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

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

16 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1) + 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 2 (u^{2} + 1) d u

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

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

\int 1 d u = u

= 8 \cdot 2 (\frac{u ^{3}}{3} + u)

Substitute back

u = x - 1

= 8 \cdot 2 (\frac{( x - 1 ) ^{3}}{3} + x - 1)

Simplify

8 \cdot 2 (\frac{( x - 1 ) ^{3}}{3} + x - 1) : 16 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1)

= 16 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1)

Add a constant to the solution

= 16 (\frac{1}{3} (x - 1)^{\frac{3}{2}} + x - 1) + C

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - \frac{25}{3}

\int \frac{( x + 2 )}{x} d x

\frac{d}{d x} (a^{x} ln (x))

11 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0

f (t) = 4 t^{2} - t

What is the integral of (8x)/(sqrt(x-1)) ?
The integral of (8x)/(sqrt(x-1)) is 16(1/3 (x-1)^{3/2}+sqrt(x-1))+C