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if the substitution u=Sqrt(x+1) then the integral from 0 to 3 [(x+2)(sqrt(x+1))
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\[\int_0^2(x+2)\sqrt{x+1}~dx\] Using the substitution \(u=\sqrt{x+1}\), you find that \(du=\dfrac{1}{2\sqrt{x+1}}~dx\), or \(2u~du=dx\). Now, you also have \(u^2=x+1\), wich gives you \(x+2=u^2+1\). When \(x=0\), you have \(u=\sqrt{0+1}=1\). When \(x=3\), \(u=\sqrt{3+1}=2\), so the integral becomes \[\int_1^2(u^2+1)~u~(2u~du)=2\int_1^2u^2\left(u^2+1\right)~du\]
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