How do you integrate by substitution $\int u^{2} \sqrt{u^{3} + 2} d u$ $int u^2sqrt(u^3+2)du$ ?

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Answer 1 · 2016-11-17T06:20:25

The answer is $= \frac{2}{9} {(u^{3} + 2)}^{\frac{3}{2}} + C$ $=2/9(u^3+2)^(3/2) +C$

We use $\int v^{n} d v = \frac{v^{n + 1}}{n + 1} + C (n \neq - 1)$ $intv^ndv=v^(n+1)/(n+1)+C (n!=-1)$

Let's do the substitution,

$x = u^{3} + 2$ $x=u^3+2$

then, $d x = 3 u^{2} d u$ $dx=3u^2du$ $\Rightarrow$ $=>$ $u^{2} d u = \frac{d x}{3}$ $u^2du=dx/3$

Therefore, $\int u^{2} \sqrt{u^{3} + 2} d u = \int \frac{\sqrt{x} d x}{3}$ $intu^2sqrt(u^3+2)du=int(sqrtxdx)/3$

$\int \frac{x^{\frac{1}{2}} d x}{3} = \frac{x^{\frac{3}{2}}}{3 \cdot \frac{3}{2}} + C$ $int(x^(1/2)dx)/3=x^(3/2)/(3*3/2)+C$

$= 2 \frac{x^{\frac{3}{2}}}{9} + C$ $=2x^(3/2)/9+C$

$\int u^{2} \sqrt{u^{3} + 2} d u = \frac{2}{9} {(u^{3} + 2)}^{\frac{3}{2}} + C$ $intu^2sqrt(u^3+2)du=2/9(u^3+2)^(3/2) +C$

How do you integrate by substitution int u^2sqrt(u^3+2)du∫u2√u3+2duint u^2sqrt(u^3+2)du?