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\int t\sqrt{t^2-4}dt

Integrate t(t^2-4)^0.5

Answer

$\frac{1}{3}\sqrt{\left(t^2-4\right)^{3}}+C_0$

Step-by-step explanation

Problem

$\int t\sqrt{t^2-4}dt$
1

Solve the integral $\int\sqrt{t^2-4}tdt$ applying u-substitution. Let $u$ and $du$ be

$\begin{matrix}u=t^2-4 \\ du=2tdt\end{matrix}$

Unlock this step-by-step solution!

Answer

$\frac{1}{3}\sqrt{\left(t^2-4\right)^{3}}+C_0$
$\int t\sqrt{t^2-4}dt$

Main topic:

Integration by substitution

Used formulas:

3. See formulas

Time to solve it:

~ 2.52 seconds