De digitale vraagbaak voor het wiskundeonderwijs

home |  vandaag |  gisteren |  bijzonder |  gastenboek |  wie is wie? |  verhalen |  contact

HOME

samengevat
vragen bekijken
een vraag stellen
hulpjes
zoeken
FAQ
links
twitter
boeken
help

inloggen

colofon

  \require{AMSmath}


Uitgewerkt

a.
$
\eqalign{
  & \int {{1 \over {1 + \sqrt x }}} \,\,dx =   \cr
  & \int {{{2\sqrt x } \over {1 + \sqrt x }}} \,\,{1 \over {2\sqrt x }}dx =   \cr
  & \int {{{2\sqrt x } \over {1 + \sqrt x }}} \,\,d\left( {\sqrt x } \right) =   \cr
  & \int {{{2u} \over {1 + u}}} \,\,du =   \cr
  & \int {2 - {2 \over {u + 1}}\,\,du = }   \cr
  & 2u - 2\ln (u + 1) + C =   \cr
  & 2\sqrt x  - 2\ln \left( {\sqrt x  + 1} \right) + C \cr}
$

b.
$
\eqalign{
  & \int {{{2x + 1} \over {\sqrt {x + 1} }}\,\,dx}  =   \cr
  & \int {2\left( {2x + 1} \right) \cdot {1 \over {2\sqrt {x + 1} }}\,\,dx}  =   \cr
  & \int {2\left( {2x + 1} \right) \cdot d\left( {\sqrt {x + 1} } \right)}  =   \cr
  & \int {2\left( {2\left( {u^2  - 1} \right) + 1} \right) \cdot du}  =   \cr
  & \int {4u^2  - 2 \cdot du}  =   \cr
  & {4 \over 3}u^3  - 2u + C  \cr
  & {4 \over 3}\left( {\sqrt {x + 1} } \right)^3  - 2\sqrt {x + 1}  + C  \cr
  & \sqrt {x + 1} \left( {{4 \over 3}\left( {\sqrt {x + 1} } \right)^2  - 2} \right) + C  \cr
  & \sqrt {x + 1} \left( {{4 \over 3}\left( {x + 1} \right) - 2} \right) + C  \cr
  & \sqrt {x + 1} \left( {{4 \over 3}x - {2 \over 3}} \right) + C  \cr
  & {2 \over 3}\sqrt {x + 1} \left( {2x - 1} \right) + C \cr}
$

c.
$
\eqalign{
  & \int {{{\tan (\ln (\sqrt x ))} \over x}} \,dx =   \cr
  & \int {\tan \left( {{1 \over 2}\ln (x)} \right)}  \cdot {1 \over x}\,dx =   \cr
  & \int {\tan \left( {{1 \over 2}\ln (x)} \right)} \,d\left( {\ln (x)} \right) =   \cr
  &  \downarrow u = \ln (x)  \cr
  & \int {\tan \left( {{1 \over 2}u} \right)} \,du =   \cr
  & \int {{{\sin \left( {{1 \over 2}u} \right)} \over {\cos \left( {{1 \over 2}u} \right)}}} \,du =   \cr
  & \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}}  \cdot  - {1 \over 2}\sin \left( {{1 \over 2}u} \right)\,du =   \cr
  & \int {{{ - 2} \over {\cos \left( {{1 \over 2}u} \right)}}}  \cdot \,d\left( {\cos \left( {{1 \over 2}u} \right)} \right) =   \cr
  &  \downarrow v = \cos \left( {{1 \over 2}u} \right)  \cr
  & \int {{{ - 2} \over v}}  \cdot \,dv =   \cr
  &  - 2\ln (v) + C =   \cr
  &  - 2\ln \left( {\cos \left( {{1 \over 2}u} \right)} \right) + C =   \cr
  &  - 2\ln \left( {\cos \left( {{1 \over 2}\ln (x)} \right)} \right) + C  \cr
  &  - 2\ln \left( {\cos \left( {\ln (\sqrt x )} \right)} \right) + C \cr}
$


home |  vandaag |  bijzonder |  gastenboek |  statistieken |  wie is wie? |  verhalen |  colofon

©2001-2024 WisFaq - versie 3