What is the integral of (x+1)sqrt(2-x) ?

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

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integral of (x+1)sqrt(2-x)

\int (x + 1) 2 - x d x

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

Solution steps

\int (x + 1) 2 - x d x

Apply u-substitution

: \int - 2 u^{2} (- u^{2} + 3) d u

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

Take the constant out:

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

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

Expand

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

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

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

\int u^{4} d u = \frac{u ^{5}}{5}

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

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

Substitute back

u = 2 - x

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

Simplify

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

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

Add a constant to the solution

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

7 sin (x), \frac{π}{4}

x d y - y d x + (y^{2} - 1) d y = 0

\frac{d}{d x} ((sin (x) + cos (x))^{2})

\frac{7 y ^{2} - t ^{2}}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

x \to 1 lim (\frac{4}{x + 3})

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