How to $\int 3t \sqrt {t^2+1} dt$

Question

$$\int 3t \sqrt {t^2+1} dt$$

My book has done

$$\int_0^2 3t (1 + t^2)^{\frac{1}{2}}dt = \left[(1 + t^2)^{\frac{3}{2}}\right]_0^2.$$

They have done so simple way, but how is this correct? I have been taught that when $t$ (inside bracket) has a power greater than $1$, this isn't possible. I used the substitution $t := \tan x$, but I can't complete it.

Answer 1

Substitute $u = t^2 + 1$. So $\frac{du}{dt} = 2t$. $$ \int 3t \sqrt {t^2+1} dt = \frac{3}{2}\int \sqrt{u} \,du = \frac{3}{2} \times \frac{u\sqrt{u}}{\frac{3}{2}}+C=(1+t^2)^{\frac{3}{2}}+C. $$

Answer 2

Using the substitution $u = 1 + t^2$, $du = 2t \,dt$, gives

$$\int 3t \sqrt{1 + t^2} dt = \frac{3}{2} \int \sqrt{u} \,du = u^{3/2} + C = (1 + t^2)^{\frac{3}{2}} + C,$$ and evaluating at the limits gives the identity.

Note that your substitution $t := \tan x$ works as well: We have $$\int 3t \sqrt{1 + t^2} dt = 3\int \tan t \sec t \cdot \sec^2 t \,dt,$$ and we can substitute $v = \sec t$, $dv = \sec t \tan t$, but note that up to an overall power this achieves the same effect that the single substitution $u$ does.

Answer 3

Hints:

$$ \int 3t \sqrt {t^2+1} dt = \int \frac32 \sqrt {t^2+1} dt^2$$