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

The integral of (x+1)sqrt(3-x) is 2/5 (3-x)^{5/2}-8/3 (3-x)^{3/2}+C

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integral of (x+1)sqrt(3-x)

\int (x + 1) 3 - x d x

\frac{2}{5} (3 - x)^{\frac{5}{2}} - \frac{8}{3} (3 - x)^{\frac{3}{2}} + C

Solution steps

\int (x + 1) 3 - x d x

Apply u-substitution

: \int - u (- u + 4) d u

= \int - u (- u + 4) d u

Take the constant out:

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

= - \int u (- u + 4) d u

Expand

u (- u + 4) : - u^{\frac{3}{2}} + 4 u

= - \int - u^{\frac{3}{2}} + 4 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

= - (- \int u^{\frac{3}{2}} d u + \int 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = 3 - x

= - (- \frac{2}{5} (3 - x)^{\frac{5}{2}} + \frac{8}{3} (3 - x)^{\frac{3}{2}})

Simplify

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

Add a constant to the solution

= \frac{2}{5} (3 - x)^{\frac{5}{2}} - \frac{8}{3} (3 - x)^{\frac{3}{2}} + C

\frac{d}{d L} ((K (L) L)^{\frac{1}{2}})

\frac{s}{s + 5}

f (x) = 4 x^{3} - x^{2} - 7, at x = - 1

\int 8 tan (x) sec^{3} (x) d x

y = (x^{2} + 2)^{2} (x^{4} + 4)^{4}

What is the integral of (x+1)sqrt(3-x) ?
The integral of (x+1)sqrt(3-x) is 2/5 (3-x)^{5/2}-8/3 (3-x)^{3/2}+C