Solved dis hw3 ex5

1 files changed, 7 insertions, 2 deletions

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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