What is the integral of (x+1)sqrt(7-x) ?

The integral of (x+1)sqrt(7-x) is 2/5 (7-x)^{5/2}-16/3 (7-x)^{3/2}+C

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integral of (x+1)sqrt(7-x)

\int (x + 1) 7 - x d x

\frac{2}{5} (7 - x)^{\frac{5}{2}} - \frac{16}{3} (7 - x)^{\frac{3}{2}} + C

Solution steps

\int (x + 1) 7 - x d x

Apply u-substitution

: \int - u (- u + 8) d u

= \int - u (- u + 8) d u

Take the constant out:

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

= - \int u (- u + 8) d u

Expand

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

= - \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

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

Substitute back

u = 7 - x

= - (- \frac{2}{5} (7 - x)^{\frac{5}{2}} + \frac{16}{3} (7 - x)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (7 - x)^{\frac{5}{2}} - \frac{16}{3} (7 - x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (7 - x)^{\frac{5}{2}} - \frac{16}{3} (7 - x)^{\frac{3}{2}} + C

\int_{0}^{1} t^{3} e^{- 2 t} d t

x \to 0 lim (\frac{e ^{x} - 1}{x ^{2}})

y = x^{2} - 12, y = 4 x

\frac{d y}{d x} (2 + x) = 3 y

x \to 1 lim (\frac{x - 1}{3})

What is the integral of (x+1)sqrt(7-x) ?
The integral of (x+1)sqrt(7-x) is 2/5 (7-x)^{5/2}-16/3 (7-x)^{3/2}+C