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

The integral from 1 to 2 of (sqrt(x)+1)^2 is 5/2+(2(4sqrt(2)-2))/3

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integral from 1 to 2 of (sqrt(x)+1)^2

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

\frac{5}{2} + \frac{2 ( 4 2 - 2 )}{3}

Decimal

4.93790 \dots

Solution steps

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

Expand

(x + 1)^{2} : x + 2 x + 1

= \int_{1}^{2} x + 2 x + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{2} x d x + \int_{1}^{2} 2 x d x + \int_{1}^{2} 1 d x

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} 2 x d x = \frac{2 ( 4 2 - 2 )}{3}

\int_{1}^{2} 1 d x = 1

= \frac{3}{2} + \frac{2 ( 4 2 - 2 )}{3} + 1

Combine the fractions

1 + \frac{3}{2} : \frac{5}{2}

= \frac{5}{2} + \frac{2 ( 4 2 - 2 )}{3}

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What is the integral from 1 to 2 of (sqrt(x)+1)^2 ?
The integral from 1 to 2 of (sqrt(x)+1)^2 is 5/2+(2(4sqrt(2)-2))/3