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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int 2 u^{2} (u^{2} + 1) d u

= \int 2 u^{2} (u^{2} + 1) 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} + 1) d u

Expand

u^{2} (u^{2} + 1) : u^{4} + u^{2}

= 2 \cdot \int u^{4} + 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 u^{2} d u)

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

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

y^{^{''}} = y + 4 t^{2}

x \to 5 lim (- 3 x)

y = 7 e^{- 6 x}, (0, 7)

\frac{d y}{d x} = 9 x - \frac{y}{x}

y^{^{''}} - 2 y^{^{'}} = 4 t, y (0) = 2, y^{^{'}} (0) = 0

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