integrals are easy actually????

src/exercise/1.typ

@@ -21,7 +21,7 @@
   &= ((2024 (1+n+n^2))/n^2)/(1+2023/n) \
   &= (2024 (1/n^2+1/n+1/))/(1+2023/n) \
   &= (2024/n^2+2024/n+2024)/(1+2023/n) \
-  
+
   => lim_(n -> infinity) a_n &= lim_(n -> infinity) (2024/n^2+2024/n+2024)/(1+2023/n) \
   &= lim_(n -> infinity) 2024/1 = 2024 checkmark \
 $
@@ -40,14 +40,14 @@ $
   &= b_1/(sqrt(1) + 1) = b_1/2 checkmark
 $
 
-+ $ 
++ $
   a_n &= (n^4-2)/(n^2+4) + (n^3(3-n^2)) / (n^3+1) \
   &= ((n^4-2)(n^3+1) + (n^3(3-n^2))(n^2+4)) / ((n^2+4)(n^3+1)) \
   &= (n^7+n^4-2n^3-2 + 3n^5+12n^3-n^7-4n^5) / ((n^2+4)(n^3+1)) \
   &= (-n^5+n^4+10n^3-2) / (n^5+n^2+4n^3+4) \
   &= (-1+1/n+10/(n^2)-2/(n^5)) / (1+1/(n^3)+4/(n^2)+4/(n^5)) \
 
-  => lim_(n -> infinity) a_n 
+  => lim_(n -> infinity) a_n
   &= lim_(n -> infinity) (-1+1/n+10/(n^2)-2/(n^5)) / (1+1/(n^3)+4/(n^2)+4/(n^5)) \
   &= lim_(n -> infinity) -1/1 = 1 checkmark
 $
@@ -72,7 +72,7 @@ $
 + $
   A &= sum_(n=1)^infinity (2^n n!)/(n^n) \
   a_n &= (2^n n!)/(n^n) #text("Ratio Test") \
-  => lim_(n -> infinity) a_n 
+  => lim_(n -> infinity) a_n
   &= lim_(n -> infinity) abs((a_(n+1))/(a_n)) \
   &= lim_(n -> infinity) abs( ( (2^(n+1) (n+1)!)/((n+1)^(n+1)) )/( (2^n n!)/(n^n) )) \
   &= lim_(n -> infinity) abs( ( 2^(n+1) dot (n+1)! dot n^n )/( 2^n dot n! dot (n+1)^(n+1) )) \
@@ -88,7 +88,7 @@ $
 
 + $
   a_n &= 1/(n^n) #text("Root Test") \
-  root(n, abs(1/(n^n))) &= 1/n -> 0 < 1 \ 
+  root(n, abs(1/(n^n))) &= 1/n -> 0 < 1 \
   &=> a_n #text(" diverges") checkmark
 $
 
@@ -109,6 +109,7 @@ $
 
 #set enum(numbering: "(a)")
 + $
+  f(x) &= ln((e^x) / (1+e^x)) \
   e(x) = e^x wide a(x) &= (e^x)/(1+e^x) wide b(x) = 1+e^x \
 
   ln: RR^+ |-> RR wide e&: RR |-> RR^+ wide => f: RR |-> RR <=> DD_f = RR \
@@ -124,3 +125,47 @@ $
   &= (1+e^x)/([1+e^x]^2) \
   &= 1/(1+e^x) checkmark
 $
++ $
+  D_f &= RR \\ {x = 1}\
+  f(x) &= ((x+1)/(x-1))^2 \
+  a(x) = x+1 wide b(x) = x-1 quad&quad c(x) = a(x)/b(x) wide d(x) = x^2 \
+  a'(x) = 1 wide b'(x) &= 1 wide d'(x) = 2x \
+  c'(x) &= (b' dot a - b dot a')/(a^2) \
+  &= (x-1 - (x+1))/((x-1)^2) \
+  &= (-2)/((x-1)^2) \
+  f'(x) &= d'(c(x)) dot c'(x) \
+  &= 2 dot (x+1)/(x-1) dot (-2)/((x-1)^2) \
+  &= -4 dot (x+1)/((x-1)^3) checkmark
+$
+
+#pagebreak()
+=== Exercise 4 @Exercise[1, 4]
+#block(
+  fill: luma(230),
+  inset: 8pt,
+  radius: 4pt,
+[
+  Calculate the following integrals using substitution:
+
+  #set enum(numbering: "(a)")
+  + $ integral (2x+7)/(x^2+7x+3) d x $
+  + $ integral (cos(ln(x)))/(x) d x $
+])
+
+#set enum(numbering: "(a)")
++ $
+  A &= integral (2x+7)/(x^2+7x+3) d x \
+  &= integral (a'(x))/(a(x)) d x \
+  &= integral (a'(x))/(a(x)) d x \
+  &= integral 1/(a(x)) dot a'(x) d x; wide d u = a'(x) d x \
+  &= integral 1/u d u \
+  &= ln(u) = ln(x^2+7x+3) checkmark
+$
++ $
+  A &= integral (cos(ln(x)))/(x) d x \
+  &= integral cos(ln(x)) dot 1/x d x \
+  &= integral cos(u) dot u' d x \
+  &= integral cos(u) dot d u \
+  &= sin(u) \
+  &= sin(ln(x)) \
+$