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diff --git a/Year 1/Calculus/Homeworks/Homework3.tex b/Year 1/Calculus/Homeworks/Homework3.tex index 49ae9ec..69ad5f1 100644 --- a/Year 1/Calculus/Homeworks/Homework3.tex +++ b/Year 1/Calculus/Homeworks/Homework3.tex @@ -66,7 +66,7 @@ Да се докаже, че за всяко $x < 0$ е изпълнено, че \begin{align*} - \frac{1}{2(1 + \sqrt{2})}\arctg \frac{x^2 - 1}{2x} + \ln \sqrt{x^2 + 1} - \frac{(\sqrt{2} - 1)\pi}{8} \geq \ln \left( 4 - 2 \sqrt{2} \right) + \frac{1}{2(1 + \sqrt{2})}\arctg \frac{x^2 - 1}{2x} + \ln \sqrt{x^2 + 1} - \frac{(\sqrt{2} - 1)\pi}{8} \geq \frac{1}{2} \ln \left( 4 - 2 \sqrt{2} \right) \end{align*} \paragraph{Решение:} [TODO] @@ -132,6 +132,11 @@ \int x \arctg(x^2) \D x \end{align*} -\paragraph{Решение:} [TODO] +\paragraph{Решение:} +\begin{align*} + & \int x \arctg(x^2) \D x = \int \arctg(x^2) \D \left( \frac{x^2}{2} \right) \xlongequal{\text{По части}} \frac{x^2}{2}\arctg(x^2) - \int \frac{x^2}{2} \D \arctg(x^2) \\ + & \qquad \int \frac{x^2}{2} \D \arctg(x^2) = \int \frac{x^2}{\cancel{2}} \frac{1}{x^4+1} \cancel{2}x \D x = \int \frac{1}{x^4+1} \cdot \frac{4}{4} x^3 \D x = \frac{1}{4} \int \frac{1}{x^4+1} d(x^4) = \frac{1}{4} \ln (x^4 + 1) + c \\ + & \implies \int x \arctg(x^2) \D x = \frac{x^2}{2}\arctg(x^2) - \frac{1}{4} \ln (x^4 + 1) + c +\end{align*} \end{document}
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