What is the integral from 0 to 1 of $x \cdot \sqrt{x^{2} + 8}$ $x*sqrt(x^2+8)$ ?

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What is the integral from 0 to 1 of $x \cdot \sqrt{x^{2} + 8}$ $x*sqrt(x^2+8)$ ?

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Answer 1 · 2015-05-06T22:06:56

Try this:
$\int x \sqrt{x^{2} + 8} d x =$ $intxsqrt(x^2+8)dx=$
But $d [x^{2} + 8] = 2 x d x$ $d[x^2+8]=2xdx$
So:
$\frac{1}{2} \int \sqrt{x^{2} + 8} d [x^{2} + 8] =$ $1/2intsqrt(x^2+8)d[x^2+8]=$
$= \frac{1}{2} \int {(x^{2} + 8)}^{\frac{1}{2}} d [x^{2} + 8] = \frac{1}{2} \frac{{(x^{2} + 8)}^{\frac{3}{2}}}{\frac{3}{2}}$ $=1/2int(x^2+8)^(1/2)d[x^2+8]=1/2((x^2+8)^(3/2))/(3/2)$
Between $0$ $0$ and $1$ $1$ :
$= \frac{1}{3} {(x^{2} + 8)}^{\frac{3}{2}} {∣ ∣ ∣}_{0}^{1}$ $=1/3(x^2+8)^(3/2)|_0^1$ = $9 - \frac{8}{3} \sqrt{8} = 9 - \frac{16}{3} \sqrt{2}$ $9-8/3sqrt(8)=9-16/3sqrt(2)$

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What is the integral from 0 to 1 of $x \cdot \sqrt{x^{2} + 8}$ $x*sqrt(x^2+8)$ ?

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What is the integral from 0 to 1 of $x \cdot \sqrt{x^{2} + 8}$ $x*sqrt(x^2+8)$ ?