A sad jedan specijalan slucaj prethodnog:
![\[\begin{array}{l}
T3. \to \int {{x^{ - 1}}} dx = \ln x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {{x^{ - 1}}} dx = \left| {{x^{ - 1}} = {e^{ - \ln x}}} \right| = \int {{e^{ - \ln x}}dx} = \left| \begin{array}{l}
- \ln x = t \\
- \frac{{dx}}{x} = dt \\
dx = - xdt = - {e^{ - t}}dt \\
\ln x = - t \to x = {e^{ - t}} \\
\end{array} \right| = \int {{e^t}( - {e^{ - t}})dt} = - \int {dt} = - t = - ( - \ln x) = \ln x + C \\
\end{array}\]](https://static.elitesecurity.org/tex/7afb65255755a1ad1c523520c3498060.png)
Integral - malo neobican
Zaključano
Zakačeno
Kad znamo T3 resavamo i sledece:
![\[\begin{array}{l}
T9. \to \int {\frac{{dx}}{{{{\sin }^2}x}}} = - {\mathop{\rm ctg}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{{{\sin }^2}x}}} = \left| \begin{array}{l}
{\mathop{\rm tg}\nolimits} x = t \\
x = {\mathop{\rm arctg}\nolimits} t \\
dx = \frac{1}{{1 + {t^2}}}dt \\
\sin x = \frac{t}{{\sqrt {1 + {t^2}} }} \\
\end{array} \right| = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{{{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}dt} = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{\frac{{{t^2}}}{{1 + {t^2}}}}}dt} = \int {\frac{{dt}}{{{t^2}}}} = \frac{{{t^{ - 1}}}}{{ - 1}} = - \frac{1}{t} = - \frac{1}{{{\mathop{\rm tg}\nolimits} x}} = - {\mathop{\rm ctg}\nolimits} x \\
\end{array}\]](https://static.elitesecurity.org/tex/2d8256a3cd0057c2fe93cfaf60ea05a2.png)
![\[\begin{array}{l}
T15. \to \int {\frac{{dx}}{{{{\sinh }^2}x}}} = - {\mathop{\rm ctgh}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{{{\sinh }^2}x}}} = \left| \begin{array}{l}
{\mathop{\rm tgh}\nolimits} x = t \\
x = {\mathop{\rm arctgh}\nolimits} t \\
dx = \frac{1}{{1 - {t^2}}}dt \\
\sinh x = \frac{t}{{\sqrt {1 - {t^2}} }} \\
\end{array} \right| = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{{{\left( {\frac{t}{{\sqrt {1 - {t^2}} }}} \right)}^2}}}dt} = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{\frac{{{t^2}}}{{1 - {t^2}}}}}dt} = \int {\frac{{dt}}{{{t^2}}}} = \frac{{{t^{ - 1}}}}{{ - 1}} = - \frac{1}{t} = - \frac{1}{{{\mathop{\rm tgh}\nolimits} x}} = - {\mathop{\rm ctgh}\nolimits} x \\
\end{array}\]](https://static.elitesecurity.org/tex/c00630c7cd0a9c65a439d723afbc86a4.png)
![\[\begin{array}{l}
T16. \to \int {\frac{{dx}}{{1 - {x^2}}}} = \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| + C = {\mathop{\rm arctgh}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{1 - {x^2}}}} = \int {\frac{1}{{(1 - x)(1 + x)}}dx} = \frac{1}{2}\int {\frac{{2 + x - x}}{{(1 - x)(1 + x)}}dx} = \frac{1}{2}\int {\frac{{(1 + x) + (1 - x)}}{{(1 - x)(1 + x)}}dx = } \\
= \frac{1}{2}\int {\frac{{(1 + x)}}{{(1 - x)(1 + x)}}dx} + \frac{1}{2}\int {\frac{{(1 - x)}}{{(1 - x)(1 + x)}}dx = } \frac{1}{2}\left( {\int {\frac{1}{{1 - x}}dx} + \int {\frac{1}{{1 + x}}dx} } \right) = \left| {\begin{array}{*{20}{c}}
\begin{array}{l}
1 - x = t \\
- dx = dt \\
\end{array} & \begin{array}{l}
1 + x = u \\
dx = du \\
\end{array} \\
\end{array}} \right| = \\
= \frac{1}{2}\left( {\int {\frac{1}{t}( - dt)} + \int {\frac{1}{u}du} } \right) = \frac{1}{2}\left( { - \ln t + \ln u} \right) = \frac{1}{2}\left( { - \ln \left| {1 - x} \right| + \ln \left| {1 + x} \right|} \right) = \frac{1}{2}\left( {\ln \left| {1 + x} \right| - \ln \left| {1 - x} \right|} \right) = \\
= \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| \\
\end{array}\]](https://static.elitesecurity.org/tex/fa3103353b30369aeb741806ace25f68.png)
![\[\begin{array}{l}
T17. \to \int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C = - {\mathop{\rm arcctgh}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{{x^2} - 1}}} = \frac{1}{2}\int {\frac{{(x + 1) - (x - 1)}}{{(x - 1)(x + 1)}}dx} = \frac{1}{2}\int {\frac{{dx}}{{x - 1}}} - \frac{1}{2}\int {\frac{{dx}}{{x + 1}}} = \left| {\begin{array}{*{20}{c}}
\begin{array}{l}
x - 1 = t \\
dx = dt \\
\end{array} & \begin{array}{l}
x + 1 = u \\
dx = du \\
\end{array} \\
\end{array}} \right| = \frac{1}{2}\left( {\int {\frac{{dt}}{t}} - \int {\frac{{du}}{u}} } \right) = \\
= \frac{1}{2}\left( {\ln t - \ln u} \right) = \frac{1}{2}\left( {\ln \left| {x - 1} \right| - \ln \left| {x + 1} \right|} \right) = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| \\
\end{array}\]](https://static.elitesecurity.org/tex/e6c54e943175ef26e72215b92d3337d1.png)
Za T18 i T19 moramo takodje upotrebiti malo zonglerskog znanja ali ovog puta sa hiberbolicnim funkcijama 
![\[\begin{array}{l}
T18. \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = \ln \left( {x + \sqrt {{x^2} + 1} } \right) + C = {\mathop{\rm arcsinh}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = \left| \begin{array}{l}
x = \sinh t \\
dx = \cosh t \\
{\mathop{\rm arcsinh}\nolimits} x = t \\
\end{array} \right| = \int {\frac{{\cosh t}}{{\sqrt {{{\sinh }^2}t + 1} }}dx} = \left| {{{\cosh }^2}t - {{\sinh }^2}t = 1} \right| = \int {\frac{{\cosh t}}{{\cosh t}}dx} = \int {dt} = t = {\mathop{\rm arcsinh}\nolimits} x \\
\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \to \left| {x = \sinh t} \right| \to x = \frac{{{e^t} - {e^{ - t}}}}{2} \\
2x = {e^t} - {e^{ - t}}/ \cdot {e^t} \\
2x{e^t} = {e^{2t}} - 1 \\
{({e^t})^2} - 2x{e^t} - 1 = 0 \to \left| \begin{array}{l}
a{x^2} + bx + c = 0 \\
{x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\end{array} \right| \to e_{1/2}^t = \frac{{2x \pm \sqrt {4{x^2} + 4} }}{2} = x \pm \sqrt {{x^2} + 1} \\
{e^t} = x + \sqrt {{x^2} + 1} \,\,\,/\ln \to \ln {e^t} = \ln (x + \sqrt {{x^2} + 1} ) \\
t = \ln (x + \sqrt {{x^2} + 1} ) \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}} = t = \ln \left( {x + \sqrt {{x^2} + 1} } \right) \\
\end{array}\]](https://static.elitesecurity.org/tex/a52fff23419b98443b9eadff11e57b77.png)
![\[\begin{array}{l}
T19. \to \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \ln \left( {x + \sqrt {{x^2} - 1} } \right) + C = {\mathop{\rm arccosh}\nolimits} x + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = \left| \begin{array}{l}
x = \cosh t \\
dx = \sinh t \\
{\mathop{\rm arccosh}\nolimits} x = t \\
\end{array} \right| = \int {\frac{{\sinh t}}{{\sqrt {{{\cosh }^2}t - 1} }}dx} = \left| {{{\cosh }^2}t - {{\sinh }^2}t = 1} \right| = \int {\frac{{\sinh t}}{{\sinh t}}dx} = \int {dt} = t = {\mathop{\rm arccosh}\nolimits} x \\
\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2} \to \left| {x = \cosh t} \right| \to x = \frac{{{e^t} + {e^{ - t}}}}{2} \\
2x = {e^t} + {e^{ - t}}/ \cdot {e^t} \\
2x{e^t} = {e^{2t}} + 1 \\
{({e^t})^2} - 2x{e^t} + 1 = 0 \to \left| \begin{array}{l}
a{x^2} + bx + c = 0 \\
{x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
\end{array} \right| \to e_{1/2}^t = \frac{{2x \pm \sqrt {4{x^2} - 4} }}{2} = x \pm \sqrt {{x^2} - 1} \\
{e^t} = x + \sqrt {{x^2} - 1} \,\,\,/\ln \to \ln {e^t} = \ln (x + \sqrt {{x^2} - 1} ) \\
t = \ln (x + \sqrt {{x^2} - 1} ) \to \int {\frac{{dx}}{{\sqrt {{x^2} - 1} }}} = t = \ln \left( {x + \sqrt {{x^2} - 1} } \right) \\
\end{array}\]](https://static.elitesecurity.org/tex/1f1f9bd609e25d8312094bad31e14525.png)
![\[\begin{array}{l}
T5. \to \int {{a^x}} dx = \frac{{{a^x}}}{{\ln a}} + C \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {{a^x}} dx = \left| \begin{array}{l}
{a^x} = t \\
{a^x}\ln adx = dt \\
dx = \frac{{dt}}{{{a^x}\ln a}} = \frac{{dt}}{{t\ln a}} \\
\end{array} \right| = \int {t\frac{{dt}}{{t\ln a}}} = \frac{1}{{\ln a}}\int {dt} = \frac{1}{{\ln a}} \cdot t = \frac{{{a^x}}}{{\ln a}} \\
\end{array}\]](https://static.elitesecurity.org/tex/df91d6d87a618e014341bb1decee0230.png)
I malo je hvalilo da 
resim preko njega samog
![\[\begin{array}{l}
T1. \to \int {} dx = x + C\,\,\,\, \\
{\rm{Dokaz}}\,{\rm{(Proof):}} \\
\int {dx} = \int {\frac{{{x^n}}}{{{x^n}}}} dx = \int {{x^n}{x^{ - n}}} dx = \left| {\begin{array}{*{20}{c}}
{dv = {x^n}dx} & {u = {x^{ - n}}} \\
{v = \frac{{{x^{n + 1}}}}{{n + 1}}} & {du = - n{x^{ - n - 1}}dx} \\
\end{array}} \right| = {x^{ - n}}\frac{{{x^{n + 1}}}}{{n + 1}} + \frac{n}{{n + 1}}\int {{x^{n + 1}}{x^{ - n - 1}}dx} = \frac{{{x^n}x}}{{{x^n}(n + 1)}} + \frac{n}{{n + 1}}\int {dx} \\
\int {dx} - \frac{n}{{n + 1}}\int {dx} = \frac{x}{{n + 1}} \\
\left( {1 - \frac{n}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\
\left( {\frac{1}{{n + 1}}} \right)\int {dx} = \frac{x}{{n + 1}} \\
\int {dx} = x + C \\
\end{array}\]](https://static.elitesecurity.org/tex/0f059b50e6b7e6bea6b04e533a6673e0.png)
resim preko njega samog
T18 je urađen onako kako sam i rekao. Uzeo si da je t = rešenju i onda ga izdiferenciraš.
To što ga ti zoveš arcsinh(x) a ne ln(x+koren(x*x + 1)) je presipanje iz šupljeg u prazno.
Nema tu ništa od žongliranja.
Žongliranje bi bila smena tg(x) = t.
To što ga ti zoveš arcsinh(x) a ne ln(x+koren(x*x + 1)) je presipanje iz šupljeg u prazno.
Nema tu ništa od žongliranja.
Žongliranje bi bila smena tg(x) = t.
Citat:
Hm...
Ti si ovde koristio T2 koji se resio koristeci T1, a sada koristis T2 da dokazes za T1...
Ae probaj opet