Příklad 4.1

[!example] Vypočtěte $$\Large \int_0^{\frac{1}{2}} 4x\cdot\arctan({2x})\;dx$$

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

D I

+ $\arctan(2x)$ $4x$

- $\frac{2}{4x^2+1}$ $2x^2$

Zapíšeme výsledek: $\arctan(2x)\cdot{2x^2}-\int_0^{\frac{1}{2}}\frac{2x^2}{4x^2 + 1}\cdot 2x^2\;dx$

$$

\large\begin{aligned}

&= \arctan(2x)\cdot{2x^2}-\int_0^{\frac{1}{2}}\frac{4x^2}{4x^2 + 1}\;dx

\end{aligned}

$$

[!tip]+ Dělení polynomů

\[\large\begin{aligned} 4x^2 &: 4x^2 +1 = 1 - \frac{1}{4x^2+1}\\ -(4x^2 + 1) \\ -1 \\ \end{aligned}\]

\[ \large\begin{aligned} &= \arctan(2x)\cdot{2x^2}-\int_0^{\frac{1}{2}}1-\frac{1}{4x^2 + 1}\;dx \\ &= \arctan(2x)\cdot{2x^2}- x - \int_0^{\frac{1}{2}}\frac{1}{(2x)^2 + 1}\;dx \\ \end{aligned} \]

[!tip]- Substituce

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

\[\large\begin{aligned} &= \arctan(2x)\cdot{2x^2}-x + \frac{1}{2}\int_0^{1}\frac{1}{u^2 + 1}\;du \\ &= \left[\arctan(2x)\cdot{2x^2}-x + \frac{1}{2}\arctan(2x)\right]^{\frac{1}{2}}_{0} \\ &= \left(\arctan(1)\cdot{2\left(\frac{1}{2}\right)^2} - \frac{1}{2} + \frac{1}{2}\arctan(1)\right) - \left(\arctan(0)\cdot{2\cdot0^2}-0 + \frac{1}{2}\arctan(0)\right) \\ &= \boxed{\arctan(1) - \frac{1}{2}} \end{aligned}\]

	D	I
+	\(\arctan(2x)\)	\(4x\)
-	\(\frac{2}{4x^2+1}\)	\(2x^2\)
Zapíšeme výsledek: \(\arctan(2x)\cdot{2x^2}-\int_0^{\frac{1}{2}}\frac{2x^2}{4x^2 + 1}\cdot 2x^2\;dx\)
$$
\large\begin{aligned}
&= \arctan(2x)\cdot{2x^2}-\int_0^{\frac{1}{2}}\frac{4x^2}{4x^2 + 1}\;dx
\end{aligned}
$$