Sea √(1+t)=u, obtenga t=u?-1, dt=2udu
∫1/[1+√(1+t)]dt
=∫2u/(1+u)du
=2∫[(1+u)-1]/(1+u)du
= 2∫du-2∫1/(1+u)d(1+u)
=2u-2ln(1+u)+C
=2√(1+ t)-2ln[1+√(1+t)]+C
Sea √(x?+a?)=t, obtenemos x?=t?-a?, dx?=2tdt
∫√(x?+a?)/xdx
=∫x√(x?+a?)/x?dx
=[∫ √ (x?+a?)/x?dx?]/2
=[∫2t?t/(t?-a?)dt]/2
=∫ [ (t?-a?)+a?]/(t?-a?)dt
=∫dt+a?∫1/(t?-a?)dt
=t+aln[(t-a)/(t+a)]/2+C
=√(x?+a?)+aln{[√(x?+a?)- a ]/[√(x?+a?)+a]}/2+C
Sea √(1+2/x)=u, obtenga x=2/(u?-1) , dx=-4u/(u?-1)?
∫√(x?+2x)/x?dx
=∫√[4/(u?-1) ) ?+4/(u?-1)]/[4/(u?-1)?]?[-4u/(u?-1)?]du
=-∫u√ [ 4+4(u?-1)/(u?-1)?]du
=-2∫u?/(u?-1)du
=- 2 ∫[(u?-1)+1]/(u?-1)du
=-2∫du-2∫1/(u?-1)du
=ln[(1+u)/(1-u)]-2u+C
=ln[(1+√(1+2/x))/(1-√(1+ 2 /x))]-2√(1+2/x)+C
Aquí √(1+2/x)=u ha sufrido dos sustituciones
Primero sea x =1/t, obtenemos dx=-1/t?dt, obtenemos ∫√(1+2t)/tdt, y luego dejamos √(1+2t)=u, es decir, √(1+2/x )=u
Sea √(e^u+1)=t, obtenga u=ln(t?-1), du=2t/(t?-1)dt
∫1/√ (e^u+1)du
=∫1/t?2t/(t?-1)dt
=∫1/(t?- 1)dt
=ln[(t-1)/(t+1)]+C
=ln[(√(e^u+1)-1)/ (√(e^ u+1)+1)]+C
Sea x=1/t, obtenga dx=-1/t?dt
∫1/x√ (a?-b ?x?)dx
=-∫t/√(a?-b?/t?)?1/t?dt
=-∫t ?/√(a ?t?-b?)?1/t?dt
=-∫1/a√[t?-(b/a)?]dt
=-ln[ t+√(t?-b?/a?)]/a+C
=-ln[1/x+√(1/x?-b?/a?)] /a+C
=ln{ax/[a+√(a?-b?x?)]}/a+C
Sea √(1+lnx)=t , obtenga x=e ^(t?-1), dx=2te^(t?-1)
∫√(1+lnx)/xlnxdx
=∫t/ (t?-1 )e^(t?-1)?2te^(t?-1)dt
=2∫t?/(t?-1)dt
=2∫[ (t?-1)+1]/(t?-1)dt
=2∫dt+2∫1/(t?-1)dt
=2t+ ln[(t-1)/(t+1)]+C
=2√(1+lnx)+l
n[(√(1+lnx)-1)/(√(1+lnx)+1)]+C