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# Find the most general antiderivative of the function. (Check your answers by differentiation.)$g(t) = \dfrac{1 + t + t^2}{\sqrt{t}}$

## $2 t^{1 / 2}+\frac{2 t^{3 / 2}}{3}+\frac{2 t^{5 / 2}}{5}+C$

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a living has to find the emotional institute of function Toothy one Steve off the spread of beauty. So this looks a little complicated, but we have likely a simple denominator used to looking at you. Break this up, Tio. Three different fractions that will make our life easier. So also, this squared of tm and every right there at the tea to the one half and there is only plus t over tea to the one half That's a plus t squared over two to the one, huh? No, I can simplify this further. Well, one over t r t one have Khun b rearing at T to the minus one, huh? And then this is the one so one minus one half. Well, thatjust t throw one half and then two minus one half. Well, that's just one point five was just three half. And now we can take the integral of this, hon Gin DT! And so this gives us. So we had to buy the general the tea to the minus one. How? Plus one. Well, the negative one, huh? Plus one plus t one. Huh? Last one over one half of one and plus Teo to the three, huh? Plus one over three, huh? Last one. There's a posse here, and then this comes out, Tio. After simplifying it, you get t to the one, huh? Over one half and then bluff. Teo three have over degree. Have in class T to the five over five and plus See, Then you take the reciprocal that was going to give it to Tio, huh? Plus two third t three, huh? Plus to fit t seven five in the plus C. And this is our anti derivatives.

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