<s>[INST] Evaluate the integral $\int_{0}^{4}\sqrt{9t^2+t^4}dt$ 
Evaluate the integral $\displaystyle\int_{0}^{4}\sqrt{9t^2+t^4}dt$.

I know I have to substitute $u=9t^2+t^4$ and $\displaystyle\frac{du}{dt}=18t+4t^3$
But what should I do hereafter?
 [/INST] $\int{\sqrt{9t^2+t^4}}dt = \int{t\sqrt{9+t^2}}dt$
$u = 9+t^2 \implies du = 2t \cdot dt \implies t \cdot dt = \frac{1}{2}du$
$\frac{1}{2}\int{\sqrt{u}}du = \frac{1}{3}u^{\frac{3}{2}}+c = \frac{1}{3}(9+t^2)^{\frac{3}{2}}+c$
$\frac{1}{3}((9+16)^{\frac{3}{2}}-9^{\frac{3}{2}})=\frac{98}{3}$
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