What is the integral of xsqrt(1-3x) ?

The integral of xsqrt(1-3x) is 1/9 (2/5 (1-3x)^{5/2}-2/3 (1-3x)^{3/2})+C

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integral of xsqrt(1-3x)

\int x 1 - 3 x d x

\frac{1}{9} (\frac{2}{5} (1 - 3 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 3 x)^{\frac{3}{2}}) + C

Solution steps

\int x 1 - 3 x d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{9} d u

= \int \frac{u ( u - 1 )}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{9} \cdot \int u^{\frac{3}{2}} - 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}{9} (\int u^{\frac{3}{2}} d u - \int u d u)

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

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

= \frac{1}{9} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - 3 x

= \frac{1}{9} (\frac{2}{5} (1 - 3 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 3 x)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{9} (\frac{2}{5} (1 - 3 x)^{\frac{5}{2}} - \frac{2}{3} (1 - 3 x)^{\frac{3}{2}}) + C

f (x) = \frac{1}{( x - 5 ) ^{2}}, at x = - 3

\int - sin (5 x) d x

\frac{ln ( x )}{2 x}

y^{^{''}} + 8 y^{^{'}} + 18 y = 0, y (0) = 1, y^{^{'}} (0) = 2

\int_{- \infty}^{- 1} \frac{1}{x} d x

What is the integral of xsqrt(1-3x) ?
The integral of xsqrt(1-3x) is 1/9 (2/5 (1-3x)^{5/2}-2/3 (1-3x)^{3/2})+C