1. ## integration

how do i intergrate

sqrt(t^2+t^4)dt?

thanks!

2. $\displaystyle \sqrt{t^2 + t^4} = \sqrt{t^2(1 + t^2)} = \sqrt{t^2}\sqrt{1 + t^2} = t\sqrt{1 + t^2}$.

So $\displaystyle \int{\sqrt{t^2 + t^4}\,dt} = \int{t\sqrt{1 + t^2}\,dt}$ which you can solve making the substitution $\displaystyle u = 1 + t^2$.

3. is it possible to do it by inspection?

4. Originally Posted by Rine198
is it possible to do it by inspection?
If you can recognise the function that differentiates to give $\displaystyle t\sqrt{1 + t^2}$, otherwise, just use the substitution...