What is the integral of xsqrt(8-4x) ?

The integral of xsqrt(8-4x) is 4/15 (-3x-4)(-x+2)^{3/2}+C

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integral of xsqrt(8-4x)

\int x 8 - 4 x d x

\frac{4}{15} (- 3 x - 4) (- x + 2)^{\frac{3}{2}} + C

Solution steps

\int x 8 - 4 x d x

Apply u-substitution

= \int \frac{u ( u - 8 )}{16} 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 u (u - 8) d u

Expand

u (u - 8) : u^{\frac{3}{2}} - 8 u

= \frac{1}{16} \cdot \int u^{\frac{3}{2}} - 8 u 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 u^{\frac{3}{2}} d u - \int 8 u d u)

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

\int 8 u d u = \frac{16}{3} u^{\frac{3}{2}}

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

Substitute back

u = 8 - 4 x

= \frac{1}{16} (\frac{2}{5} (8 - 4 x)^{\frac{5}{2}} - \frac{16}{3} (8 - 4 x)^{\frac{3}{2}})

Simplify

\frac{1}{16} (\frac{2}{5} (8 - 4 x)^{\frac{5}{2}} - \frac{16}{3} (8 - 4 x)^{\frac{3}{2}}) : \frac{4}{15} (- 3 x - 4) (- x + 2)^{\frac{3}{2}}

= \frac{4}{15} (- 3 x - 4) (- x + 2)^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{15} (- 3 x - 4) (- x + 2)^{\frac{3}{2}} + C

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\int (2 x^{- 2} + 4) d x

What is the integral of xsqrt(8-4x) ?
The integral of xsqrt(8-4x) is 4/15 (-3x-4)(-x+2)^{3/2}+C