[breaks things] remove some packages

2 files changed, 41 insertions, 28 deletions

diff --git a/intro.tex b/intro.tex
index cf1846d..a196602 100644
--- a/intro.tex
+++ b/intro.tex
@@ -5,7 +5,7 @@
 \setlength\headheight{13.6pt}
 
 \usepackage{
-  amsmath, amsthm, amssymb, mathtools, commath,
+  amsmath, amssymb, mathtools,
   graphicx, xcolor,
   fancyhdr, hyperref, enumerate, framed
 }
@@ -43,7 +43,19 @@
     \end{cframed}
 }
 
+\newcommand{\dmquestion}[1]{
+    \begin{center} \textbf{#1} \end{center}
+}
+
 \newcommand{\br}[1]{\left( #1 \right)}
+\newcommand*{\qed}{\hfill\ensuremath{\blacksquare}}
+\newcommand*{\qedempty}{\hfill\ensuremath{\square}}
+
+\newcommand{\explain}[1]{
+    \begin{bmatrix}
+        #1
+    \end{bmatrix}
+}
 
 \newcommand{\probability}[1]{\mathrm{Pr} \left[ #1 \right]}
 \newcommand{\expected}[1]{\mathrm{E} \left[ #1 \right]}
@@ -51,6 +63,7 @@
 \newcommand{\todo}{\texttt{todo!}}
 
 \newcommand{\osmall}[1]{\overline{o}\left( #1 \right)}
+\DeclareMathOperator{\dif}{d \!}
 
 \hypersetup{colorlinks=true, linkcolor=magenta}
 
diff --git a/sol0120.tex b/sol0120.tex
index 10ecfbb..f073973 100644
--- a/sol0120.tex
+++ b/sol0120.tex
@@ -23,29 +23,29 @@
     \]
 }
     \[
-        \braced{ \frac{1}{(1 - s)x^{s - 1}} + C }' =
+        \ br{ \frac{1}{(1 - s)x^{s - 1}} + C }' =
         -\frac{0 - (1 - s)(s - 2)x^{s - 2}}{(1 - s)^2 x^{2s - 2}} =
         \frac{1}{x^s} \qed
     \]
 
 \question{Лемма 2}{
     \[
-        \int \frac{dx}{(x^2 + a^2)^2} = \frac{1}{2a^2} \braced{
+        \int \frac{dx}{(x^2 + a^2)^2} = \frac{1}{2a^2} \ br{
             \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a} 
         } + C
     \]
 }
 
     \begin{align*}
-        \braced{
-            \frac{1}{2a^2} \braced{
+        \ br{
+            \frac{1}{2a^2} \ br{
                 \frac{x}{x^2 + a^2} + \frac{1}{a}\arctan\frac{x}{a}
             }
         }' =
-        \frac{1}{2a^2} \braced{
+        \frac{1}{2a^2} \ br{
             \frac{x^2 + a^2 - 2x^2}{(x^2 + a^2)^2} + \frac{1}{x^2 + a^2}
         } =
-        \frac{1}{2a^2} \braced{
+        \frac{1}{2a^2} \ br{
             \frac{2a^2}{(x^2 + a^2)^2}
         } =
         \frac{1}{(x^2 + a^2)^2} \qed
@@ -115,7 +115,7 @@
     \begin{gather*}
         \frac{1}{8}\int \frac{x + 2}{x^2 + 2x + 2} dx +
         \frac{1}{8}\int \frac{2 - x}{x^2 - 2x + 2} dx =\\
-        =\frac{1}{16}\braced{
+        =\frac{1}{16}\ br{
             \log | x^2 + 2x + 2 | + 2\arctan(x + 1) -
             \log | x^2 - 2x + 2 | + 2\arctan(x - 1)
         } + \bar{C}
@@ -123,7 +123,7 @@
 
 \question{(seminar0113) 8.b}{
     \[
-        \int \frac{x^5 - x}{x^8 + 1}dx = \frac{\sqrt{2}}{8} \braced{
+        \int \frac{x^5 - x}{x^8 + 1}dx = \frac{\sqrt{2}}{8} \ br{
             \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
         } + \bar{C}
     \]
@@ -140,7 +140,7 @@
         \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du
     \]
     \[
-        u^4 + 1 = \braced{u^2 + \sqrt{2} u + 1}\braced{u^2 - \sqrt{2} u + 1}
+        u^4 + 1 = \ br{u^2 + \sqrt{2} u + 1}\ br{u^2 - \sqrt{2} u + 1}
     \]
     \[
         \frac{u^2 - 1}{u^4 + 1} =
@@ -172,7 +172,7 @@
         \begin{gather*}
             \int \frac{-u - \invsq}{u^2 + \sqrt{2} u + 1} du =\\
             -\int \frac{u + \invsq}{u^2 + \sqrt{2} u + 1} du =\\
-            -\frac{1}{2} \int \frac{d \braced{ u^2 + \sqrt{2} u + 1 }}{u^2 + \sqrt{2} u + 1} =\\
+            -\frac{1}{2} \int \frac{d \ br{ u^2 + \sqrt{2} u + 1 }}{u^2 + \sqrt{2} u + 1} =\\
             -\frac{1}{2} \log |u^2 + \sqrt{2} u + 1| + C_1
         \end{gather*}
     \end{minipage}
@@ -182,7 +182,7 @@
         \begin{gather*}
             \int \frac{u - \invsq}{u^2 - \sqrt{2} u + 1} du =\\
             \int \frac{u + \invsq}{u^2 - \sqrt{2} u + 1} du =\\
-            \frac{1}{2} \int \frac{d \braced{ u^2 - \sqrt{2} u + 1 }}{u^2 - \sqrt{2} u + 1} =\\
+            \frac{1}{2} \int \frac{d \ br{ u^2 - \sqrt{2} u + 1 }}{u^2 - \sqrt{2} u + 1} =\\
             \frac{1}{2} \log |u^2 - \sqrt{2} u + 1| + C_2
         \end{gather*}
         \end{tabular}
@@ -190,7 +190,7 @@
 
     \[
         \frac{1}{2} \int \frac{u^2 - 1}{u^4 + 1} du =
-        \frac{\sqrt{2}}{8} \braced{
+        \frac{\sqrt{2}}{8} \ br{
             \log |u^2 - \sqrt{2} u + 1| - \log |u^2 + \sqrt{2} u + 1|
         } + C_3
     \]
@@ -198,7 +198,7 @@
     Обратно к $x$:
     \[
         \int \frac{x^5 - x}{x^8 + 1}dx =
-        \frac{\sqrt{2}}{8} \braced{
+        \frac{\sqrt{2}}{8} \ br{
             \log |x^4 - \sqrt{2} x^2 + 1| - \log |x^4 + \sqrt{2} x^2 + 1|
         } + \bar{C}
     \]
@@ -206,7 +206,7 @@
 \clearpage
 \question{(seminar0113) 13}{
     \[
-        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \braced{
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \ br{
             -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
         } + \bar{C}
     \]
@@ -230,23 +230,23 @@
         \int -\frac{2}{5} \frac{dx}{x + 2} = -\frac{2 \log |x + 2|}{5} + C_1\\[16pt]
         \int \frac{3}{10} \frac{dx}{x + 3} = \frac{3 \log |x + 3|}{10} + C_2\\[16pt]
         \int \frac{1}{10} \frac{(x + 1) dx}{x^2 + 1} =
-            \frac{1}{10} \braced{
+            \frac{1}{10} \ br{
                 \frac{1}{2}\int \frac{2x \ dx}{x^2 + 1} + \int \frac{dx}{x^2 + 1}
             } =
-            \frac{1}{10} \braced{
+            \frac{1}{10} \ br{
                 \frac{1}{2} \log (x^2 + 1) + \arctan(x)
             } + C_3
     \end{gather*}
 
     \[
-        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \braced{
+        \int \frac{x \ dx}{(x^2 + 1)(x + 2)(x + 3)} = \frac{1}{20} \ br{
             -8 \log |x + 2| + 6 \log |x + 3| + \log (x^2 + 1) + 2\arctan x
         } + \bar{C}
     \]
 
 \question{(seminar0120) 2.4}{
     \[
-        \int \frac{dx}{x(x^2 + 1)^2} = -\frac{1}{2} \braced{
+        \int \frac{dx}{x(x^2 + 1)^2} = -\frac{1}{2} \ br{
             -\log (x^2 + 1) +
             \frac{1}{x^2 + 1} +
             2\log |x|
@@ -324,24 +324,24 @@
                 \text{Лемма 1}
             \end{bmatrix}\\[8pt]
         \int \frac{-2x + 3}{9(x^2 - x + 1)}dx &=
-            -\frac{1}{9} \braced{
+            -\frac{1}{9} \ br{
                 \int \frac{(2x - 1) dx}{x^2 - x + 1} -
-                \int \frac{2 dx}{\braced{ x - \frac{1}{2} }^2 + \frac{3}{4}}
+                \int \frac{2 dx}{\ br{ x - \frac{1}{2} }^2 + \frac{3}{4}}
             }\nonumber \\[8pt]
             &=
-            -\frac{1}{9} \braced{
+            -\frac{1}{9} \ br{
                 \log (x^2 - x + 1) -
                 \frac{4}{\sqrt{3}} \arctan \frac{2x - 1}{\sqrt{3}}
             } + C_3\\[8pt]
         \frac{1}{3} \int \frac{(1 - x)dx}{(x^2 - x + 1)^2} &=
-            -\frac{1}{6} \braced{
+            -\frac{1}{6} \ br{
                 \int \frac{(2x - 1)dx}{(x^2 - x + 1)^2} -
-                \int \frac{dx}{\braced{ \braced{ x - \frac{1}{2} }^2 + \frac{3}{4} }^2}
+                \int \frac{dx}{\ br{ \ br{ x - \frac{1}{2} }^2 + \frac{3}{4} }^2}
             }\nonumber \\[8pt]
             &=
-            -\frac{1}{6} \braced{
+            -\frac{1}{6} \ br{
                 -\frac{1}{x^2 - x + 1} +
-                \frac{2}{3} \braced{
+                \frac{2}{3} \ br{
                     \frac{x}{x^2 - x + 1} +
                     \frac{2}{\sqrt{3}} \arctan\frac{2x - 1}{\sqrt{3}}
                 }
@@ -351,7 +351,7 @@
                 \text{Лемма 2 на правую часть}
             \end{bmatrix}
             \nonumber \\[8pt]
-            &= \frac{1}{9} \braced{
+            &= \frac{1}{9} \ br{
                 \frac{2x - 1}{x^2 - x + 1} -
                 \frac{2}{\sqrt{3}} \arctan \frac{2x - 1}{\sqrt{3}}
             } + C_4
@@ -363,7 +363,7 @@
         \int \frac{dx}{9(x + 1)^2} +
         \int \frac{-2x + 3}{9(x^2 - x + 1)}dx +
         \frac{1}{3} \int \frac{(1 - x)dx}{(x^2 - x + 1)^2} =\\[16pt]
-        \frac{1}{9} \braced{
+        \frac{1}{9} \ br{
             2\log |x + 1| - \frac{1}{x + 1}
             - \log(x^2 - x + 1) + \frac{4}{\sqrt{3}}\arctan \frac{2x - 1}{\sqrt{3}}
             - \frac{2x - 1}{x^2 - x + 1} - \frac{2}{\sqrt{3}}\arctan\frac{2x - 1}{\sqrt{3}}