Question 819287
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
Note:  {{{int( x^n,dx)= x^(n+1)/(n+1) + C}}}(n &#8800;  1)  & {{{int( e^u,du)= e^u +C}}} & {{{int( 1/u,du)}}}= ln|x| + C  
Integrate the following functions using substitution method.
1) {{{int( green((x+5))^7,dx)}}}    Let u = x+5,   {{{(du)/(dx)= 1}}},  {{{(du)= dx}}}&#61614; Then: 
{{{int( green((x+5))^7,dx)}}} = {{{int( u^7,du) = u^8/8 + C = highlight((x+5)^8/8 +C)}}} 
2) {{{int( 2x(green(x^2+3))^5,dx)}}}  Let u = x^2+3, {{{(du)= 2x(dx)}}}Then:
{{{int( 2x(green(x^2+3))^5,dx)}}} = {{{int( u^5,du)= u^6/6 + C = highlight( (x^2+3)^6/6  + C)}}} 
3) {{{int( (2t+1)e^green((t^2+t)), dt )}}}   Let u = t^2+t, {{{(du)/(dt)= 2t+1}}},{{{(du)= (2t+1)dt}}}Then:
{{{int( (2t+1)e^green((t^2+t)), dt )}}} = {{{int( e^u,du)= e^u +C = highlight( e^(t^2+t)+ C)}}} 
4) {{{int( (e^x+e^(2x))/e^x ,dx) = int( 1+e^(x),dx) = highlight(x + e^x + C) }}}
5) {{{int( 3/(green(1+2x)), dx) }}}   Let u = 1 + 2x, {{{(du)/(dx)= 2}}}, {{{(du)= 2(dx)}}} Then:
{{{int( 3/(green(1+2x)), dx) }}} = {{{(2/3)(int( 1/u,du))}}}= <sup>2</sup>&frasl;<sub>3</sub>  ln|u| + C = <u><sup>2</sup>&frasl;<sub>3</sub> ln|1+2x| + C</u>