What is the integral of x/((5-4x)^{3/2)} ?

The integral of x/((5-4x)^{3/2)} is (-2x+5)/(4(-4x+5)^{1/2)}+C

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integral of x/((5-4x)^{3/2)}

\int \frac{x}{( 5 - 4 x ) ^{\frac{3}{2}}} d x

\frac{- 2 x + 5}{4 ( - 4 x + 5 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x}{( 5 - 4 x ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{u - 5}{16 u ^{\frac{3}{2}}} d u

= \int \frac{u - 5}{16 u ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{u - 5}{u ^{\frac{3}{2}}} d u

Expand

\frac{u - 5}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{1}{2}}} - \frac{5}{u ^{\frac{3}{2}}}

= \frac{1}{16} \cdot \int \frac{1}{u ^{\frac{1}{2}}} - \frac{5}{u ^{\frac{3}{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

= \frac{1}{16} (\int \frac{1}{u ^{\frac{1}{2}}} d u - \int \frac{5}{u ^{\frac{3}{2}}} d u)

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

\int \frac{5}{u ^{\frac{3}{2}}} d u = - \frac{10}{u ^{\frac{1}{2}}}

= \frac{1}{16} (2 u^{\frac{1}{2}} - (- \frac{10}{u ^{\frac{1}{2}}}))

Substitute back

u = 5 - 4 x

= \frac{1}{16} (2 (5 - 4 x)^{\frac{1}{2}} - (- \frac{10}{( 5 - 4 x ) ^{\frac{1}{2}}}))

Simplify

\frac{1}{16} (2 (5 - 4 x)^{\frac{1}{2}} - (- \frac{10}{( 5 - 4 x ) ^{\frac{1}{2}}})) : \frac{- 2 x + 5}{4 ( - 4 x + 5 ) ^{\frac{1}{2}}}

= \frac{- 2 x + 5}{4 ( - 4 x + 5 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{- 2 x + 5}{4 ( - 4 x + 5 ) ^{\frac{1}{2}}} + C

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What is the integral of x/((5-4x)^{3/2)} ?
The integral of x/((5-4x)^{3/2)} is (-2x+5)/(4(-4x+5)^{1/2)}+C