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

The integral from 0 to 2 of (4+x^2)^{3/2} is 6ln(1+2^{1/2})+14*2^{1/2}

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integral from 0 to 2 of (4+x^2)^{3/2}

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

6 ln (1 + 2^{\frac{1}{2}}) + 14 \cdot 2^{\frac{1}{2}}

Decimal

25.08723 \dots

Solution steps

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

Simplify

x^{2} : xx

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

Apply Integration By Parts

: [x (4 + x^{2})^{\frac{3}{2}} - \int 3 x^{2} (4 + x^{2})^{\frac{1}{2}} d x]_{0}^{2}

= [x (4 + x^{2})^{\frac{3}{2}} - \int 3 x^{2} (4 + x^{2})^{\frac{1}{2}} d x]_{0}^{2}

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

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

Simplify

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

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

Compute the boundaries:

6 ln (1 + 2^{\frac{1}{2}}) + 14 \cdot 2^{\frac{1}{2}}

= 6 ln (1 + 2^{\frac{1}{2}}) + 14 \cdot 2^{\frac{1}{2}}

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