What is the integral from 0 to 2 of sqrt(4+x^2)^3 ?

The integral from 0 to 2 of sqrt(4+x^2)^3 is 6ln(1+sqrt(2))+14sqrt(2)

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integral from 0 to 2 of sqrt(4+x^2)^3

\int_{0}^{2} 4 + x^{2}^{3} d x

6 ln (1 + 2) + 142

Decimal

25.08723 \dots

Solution steps

\int_{0}^{2} (4 + x^{2})^{3} d x

Simplify

(4 + x^{2})^{3} : (4 + x^{2})^{2} 4 + x^{2}

= \int_{0}^{2} (4 + x^{2})^{2} 4 + x^{2} d x

(4 + x^{2})^{2} = 4 + x^{2}

= \int_{0}^{2} (4 + x^{2}) 4 + x^{2} d x

Apply Integration By Parts

: [x (4 + x^{2}) 4 + x^{2} - \int 3 x^{2} 4 + x^{2} d x]_{0}^{2}

= [x (4 + x^{2}) 4 + x^{2} - \int 3 x^{2} 4 + x^{2} d x]_{0}^{2}

\int 3 x^{2} 4 + x^{2} d x = 48 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}})

= [x (4 + x^{2}) 4 + x^{2} - 48 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}})]_{0}^{2}

Simplify

[x (4 + x^{2}) 4 + x^{2} - 48 (- \frac{1}{32} (x 4 + x^{2} + 4 ln (\frac{x + 4 + x ^{2}}{2})) + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}})]_{0}^{2} : [x^{3} x^{2} + 4 + \frac{11}{2} x x^{2} + 4 - \frac{3}{4} x (x^{2} + 4)^{\frac{3}{2}} + 6 ln (\frac{1}{2} x + x^{2} + 4)]_{0}^{2}

= [x^{3} x^{2} + 4 + \frac{11}{2} x x^{2} + 4 - \frac{3}{4} x (x^{2} + 4)^{\frac{3}{2}} + 6 ln (\frac{1}{2} x + x^{2} + 4)]_{0}^{2}

Compute the boundaries:

6 ln (1 + 2) + 142

= 6 ln (1 + 2) + 142

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What is the integral from 0 to 2 of sqrt(4+x^2)^3 ?
The integral from 0 to 2 of sqrt(4+x^2)^3 is 6ln(1+sqrt(2))+14sqrt(2)