^2 x+2 dx Let x+2=t =t-2 =dt Now, ^2 x+2 dx = (t-2)^2 t dt = (4^2-4t+4)t^1 2 dt = (t^ 2+1 2 -4t^ 1+1 2 +4t^ 1 2 )dt = (t^ 5 2 -4t^ 3 2 +4t^ 1 2 )dt = [ t^ 5 2 +1 5 2 +1 ]-4 [ t^ 3 2 +1 3 2 +1 ]+4 [ t^ 1 2 +1 1 2 +1 ]+c =2 7 t^ 7 2 -8 5 t^ 5 2 +8 3 t^ 3 2 +C =2 7 (x+2)^7 3 -8 5 (x+2)^5 2 +8 3 (x+2)^3 2 +C

^2 x+2 dx

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$\int\text{x}^2\sqrt{\text{x}+2}\text{ dx}$

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Answer

$\int\text{x}^2\sqrt{\text{x}+2}\text{ dx}$
Let $\text{x}+2=\text{t}$
$\Rightarrow\text{x}=\text{t}-2$
$\Rightarrow\text{dx}=\text{dt}$
Now, $\int\text{x}^2\sqrt{\text{x}+2}\text{ dx}$
$=\int(\text{t}-2)^2\sqrt{\text{t}}\text{ dt}$
$=\int(4^2-4\text{t}+4)\text{t}^\frac{1}{2}\text{dt}$
$=\int\Big(\text{t}^{2+\frac{1}{2}}-4\text{t}^{1+\frac{1}{2}}+4\text{t}^{\frac{1}{2}}\Big)\text{dt}$
$=\int\Big(\text{t}^{\frac{5}{2}}-4\text{t}^{\frac{3}{2}}+4\text{t}^{\frac{1}{2}}\Big)\text{dt}$
$=\Bigg[\frac{\text{t}^{\frac{5}{2}+1}}{\frac{5}{2}+1}\Bigg]-4\Bigg[\frac{\text{t}^{\frac{3}{2}+1}}{\frac{3}{2}+1}\Bigg]+4\Bigg[\frac{\text{t}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\Bigg]+\text{c}$
$=\frac{2}{7}\text{t}^{\frac{7}{2}}-\frac{8}{5}\text{t}^{\frac{5}{2}}+\frac{8}{3}\text{t}^{\frac{3}{2}}+\text{C}$
$=\frac{2}{7}(\text{x}+2)^\frac{7}{3}-\frac{8}{5}(\text{x}+2)^\frac{5}{2}+\frac{8}{3}(\text{x}+2)^\frac{3}{2}+\text{C}$

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