What is the integral from 0 to 3 of (x+2)sqrt(x+1) ?

The integral from 0 to 3 of (x+2)sqrt(x+1) is 256/15

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integral from 0 to 3 of (x+2)sqrt(x+1)

\int_{0}^{3} (x + 2) x + 1 d x

\frac{256}{15}

Decimal Notation

17.06666 \dots

Solution steps

\int_{0}^{3} (x + 2) x + 1 d x

Apply u-substitution

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

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

Expand

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

= 2 \cdot \int_{1}^{2} 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_{1}^{2} u^{4} d u + \int_{1}^{2} u^{2} d u)

\int_{1}^{2} u^{4} d u = \frac{31}{5}

\int_{1}^{2} u^{2} d u = \frac{7}{3}

= 2 (\frac{31}{5} + \frac{7}{3})

Simplify

2 (\frac{31}{5} + \frac{7}{3}) : \frac{256}{15}

= \frac{256}{15}

\int \frac{x ^{14}}{x ^{5} - 1} d x

\frac{\partial}{\partial z} (e^{x yz})

\int (e^{3 x} - 4)^{2} d x

\frac{x}{x ^{2} + 64}

y = 3 x^{3} - x^{2} + 9, (2, 29)

What is the integral from 0 to 3 of (x+2)sqrt(x+1) ?
The integral from 0 to 3 of (x+2)sqrt(x+1) is 256/15