Příklad 4.58

[!example] Vypočtěte $$\Large \int_{0}^{1} 8x\cdot\arctan(2x)\;dx$$

\[\large \begin{align} &= \int_{0}^{1} 4x\cdot\arctan(2x)\;2dx \\ \end{align} \]

[!tip]+ Substituce

\[\large\begin{aligned} u &= 2x \\ du &= 2dx \\ \end{aligned}\]

\[\large \begin{align} &= \int_{0}^{2} 2u\cdot\arctan(u)\;2du \\ &= 2\int_{0}^{2} u\cdot\arctan(u)\;2du \\ \end{align} \]

[!tip]+ Aplikace [[Per Partes#Metoda DI|DI metody]] Sestavíme si DI tabulku:

D I

+ $\arctan(u)$ $u$

- $\frac{1}{u^2+1}$ $\frac{u^2}{2}$

Zapíšeme výsledek: $\arctan(u)\cdot\frac{u^2}{2}-\int_0^{2}\frac{1}{u^2+1}\cdot\frac{u^2}{2}\;du$

\[\large \begin{align} &= 2\left[\arctan(u)\cdot\frac{u^2}{2}\right]_{0}^{2}-2\int_0^{2}\frac{1}{u^2+1}\cdot\frac{u^2}{2}\;du \\ &= \left[\arctan(u)\cdot{u^2}\right]_{0}^{2}-\int_0^{2}\frac{u^2}{u^2+1}\;du \\ \end{align} \]

[!tip]+ Dělení polynomů

\[\large\begin{aligned} (u^2) &: (u^2+1) = 1 - \frac{1}{u^2 + 1}\\ -(u^2+1) \\ \\ -1\\ \end{aligned}\]

\[\large \begin{align} &= \left[\arctan(u)\cdot{u^2}\right]_{0}^{2}-\int_0^{2}1 - \frac{1}{u^2 + 1}\;du \\ &= \left[\arctan(u)\cdot{u^2}\right]_{0}^{2}-\int_0^{2}1\;du + \int_0^{2}\frac{1}{u^2 + 1}\;du \\ &= \left[\arctan(u)\cdot{u^2} - u\right]_{0}^{2} + \int_0^{2}\frac{1}{u^2 + 1}\;du \\ &= \left[\arctan(u)\cdot{u^2} - u + \arctan(u)\right]_{0}^{2} \\ \end{align} \]

[!info] $\arctan(0) = 0$

\[\large \begin{align} &= \left[4\arctan(2) - 2 + \arctan(2)\right] - \left[0\arctan(0) - 0 + \arctan(0)\right]\\ &= 4\arctan(2) - 2 + \arctan(2)\\ &= \boxed{5\arctan(2) - 2}\\ \end{align} \]

	D	I
+	\(\arctan(u)\)	\(u\)
-	\(\frac{1}{u^2+1}\)	\(\frac{u^2}{2}\)
Zapíšeme výsledek: \(\arctan(u)\cdot\frac{u^2}{2}-\int_0^{2}\frac{1}{u^2+1}\cdot\frac{u^2}{2}\;du\)