Найти интеграл: \( \int \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} dx \)

Решение

\( \int \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} dx = \begin{vmatrix}
t = \sqrt{2x-1}\\
t' = \frac{1}{2\sqrt{2x-1}} \cdot 2 = \frac{1}{\sqrt{2x-1}}\\
dx = \sqrt{2x-1}\\
\end{vmatrix} = \int \mathop{\mathrm{ arctg {t}}} \cdot \sqrt{2x-1} dt = \)
\(= \int \mathop{\mathrm{ arctg {t}}} \cdot t dt = \begin{vmatrix}\\
u = \mathop{\mathrm{ arctg {t}}}\\
dv = t dt\\
du = v'(t) dt = \dfrac{1}{1+t^2} dt\\
v (t) = \dfrac{t^2}{2}\\
\end{vmatrix} = \mathop{\mathrm{ arctg {t}}} \cdot \frac{t^2}{2} - \int \frac{t^2}{2}\cdot \frac{1}{1+t^2} dt = \mathop{\mathrm{ arctg {t}}} \cdot \frac{t^2}{2} - \int \frac{t^2}{2(1+t^2)} dt =\)

\(= \mathop{\mathrm{ arctg {t}}} \cdot \dfrac{t^2}{2} - \dfrac{1}{2} \int \dfrac{t^2}{1+t^2} dt =
\mathop{\mathrm{ arctg {t}}} \cdot \dfrac{t^2}{2} - \dfrac{1}{2} \int \dfrac{t^2+1-1}{1+t^2} dt =
\mathop{\mathrm{ arctg {t}}} \cdot \dfrac{t^2}{2} - \dfrac{1}{2} \int \dfrac{t^2+1}{1+t^2} - \dfrac{1}{1+t^2} dt=\)

\(= \mathop{\mathrm{ arctg {t}}} \cdot \dfrac{t^2}{2} - \dfrac{1}{2} \cdot (\int \frac{t^2+1}{1+t^2} dt - \int \dfrac{1}{1+t^2} dt) = \mathop{\mathrm{ arctg {t}}} \cdot \dfrac{t^2}{2} - \dfrac{1}{2} \cdot (t - \mathop{\mathrm{ arctg {t}}} ) + C=\)

\(= \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} \cdot \dfrac{(\sqrt{2x-1})^2}{2} - \dfrac{1}{2}\cdot (\sqrt{2x-1} - \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} )+ C = \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} \cdot x - \dfrac{1}{2} \cdot \sqrt{2x-1} + C
\)

Ответ: \( \mathop{\mathrm{ arctg {\sqrt{2x-1}}}} \cdot x - \dfrac{1}{2} \cdot \sqrt{2x-1} + C \).