What is the integral of x/((0.5x-1)^{0.5)} ?

The integral of x/((0.5x-1)^{0.5)} is 4(0.66666…(0.5x-1)^{3/2}+2sqrt(0.5x-1))+C

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integral of x/((0.5x-1)^{0.5)}

integral

\frac{x}{( 0.5 x - 1 ) ^{0.5}}

4 (0.66666 \dots (0.5 x - 1)^{\frac{3}{2}} + 2 0.5 x - 1) + C

Solution steps

\int \frac{x}{( 0.5 x - 1 ) ^{0.5}} d x

Apply u-substitution

= \int \frac{2 ( u + 1 )}{0.5 u ^{\frac{1}{2}}} d u

Take the constant out:

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

= \frac{2}{0.5} \cdot \int \frac{u + 1}{u ^{\frac{1}{2}}} d u

Simplify

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

Expand

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

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

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

\int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}}

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

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

Substitute back

u = 0.5 x - 1

= 4 (\frac{2}{3} (0.5 x - 1)^{\frac{3}{2}} + 2 (0.5 x - 1)^{\frac{1}{2}})

Simplify

4 (\frac{2}{3} (0.5 x - 1)^{\frac{3}{2}} + 2 (0.5 x - 1)^{\frac{1}{2}}) : 4 (0.66666 \dots (0.5 x - 1)^{\frac{3}{2}} + 2 0.5 x - 1)

= 4 (0.66666 \dots (0.5 x - 1)^{\frac{3}{2}} + 2 0.5 x - 1)

Add a constant to the solution

= 4 (0.66666 \dots (0.5 x - 1)^{\frac{3}{2}} + 2 0.5 x - 1) + C

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What is the integral of x/((0.5x-1)^{0.5)} ?
The integral of x/((0.5x-1)^{0.5)} is 4(0.66666…(0.5x-1)^{3/2}+2sqrt(0.5x-1))+C