Integrate t(t^2-4)^0.5

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

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Answer

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

Step by step solution

Problem

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

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

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

Isolate $dt$ in the previous equation

$\frac{du}{2t}=dt$
3

Substituting $u$ and $dt$ in the integral

$\int\frac{\sqrt{u}}{2}du$
4

Taking the constant out of the integral

$\frac{1}{2}\int\sqrt{u}du$
5

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{1}{2}\cdot \frac{2}{3}\sqrt{u^{3}}$
6

Substitute $u$ back for it's value, $t^2-4$

$\frac{1}{3}\sqrt{\left(t^2-4\right)^{3}}$
7

Add the constant of integration

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

Answer

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

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Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.46 seconds

Views:

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