integrals are easy actually????
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@ -109,6 +109,7 @@ $
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#set enum(numbering: "(a)")
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+ $
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f(x) &= ln((e^x) / (1+e^x)) \
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e(x) = e^x wide a(x) &= (e^x)/(1+e^x) wide b(x) = 1+e^x \
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ln: RR^+ |-> RR wide e&: RR |-> RR^+ wide => f: RR |-> RR <=> DD_f = RR \
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@ -124,3 +125,47 @@ $
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&= (1+e^x)/([1+e^x]^2) \
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&= 1/(1+e^x) checkmark
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$
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+ $
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D_f &= RR \\ {x = 1}\
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f(x) &= ((x+1)/(x-1))^2 \
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a(x) = x+1 wide b(x) = x-1 quad&quad c(x) = a(x)/b(x) wide d(x) = x^2 \
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a'(x) = 1 wide b'(x) &= 1 wide d'(x) = 2x \
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c'(x) &= (b' dot a - b dot a')/(a^2) \
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&= (x-1 - (x+1))/((x-1)^2) \
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&= (-2)/((x-1)^2) \
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f'(x) &= d'(c(x)) dot c'(x) \
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&= 2 dot (x+1)/(x-1) dot (-2)/((x-1)^2) \
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&= -4 dot (x+1)/((x-1)^3) checkmark
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$
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#pagebreak()
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=== Exercise 4 @Exercise[1, 4]
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#block(
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fill: luma(230),
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inset: 8pt,
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radius: 4pt,
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[
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Calculate the following integrals using substitution:
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#set enum(numbering: "(a)")
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+ $ integral (2x+7)/(x^2+7x+3) d x $
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+ $ integral (cos(ln(x)))/(x) d x $
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])
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#set enum(numbering: "(a)")
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+ $
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A &= integral (2x+7)/(x^2+7x+3) d x \
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&= integral (a'(x))/(a(x)) d x \
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&= integral (a'(x))/(a(x)) d x \
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&= integral 1/(a(x)) dot a'(x) d x; wide d u = a'(x) d x \
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&= integral 1/u d u \
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&= ln(u) = ln(x^2+7x+3) checkmark
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$
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+ $
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A &= integral (cos(ln(x)))/(x) d x \
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&= integral cos(ln(x)) dot 1/x d x \
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&= integral cos(u) dot u' d x \
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&= integral cos(u) dot d u \
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&= sin(u) \
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&= sin(ln(x)) \
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$
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