Question #362bd

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Answer 1 · 2016-11-18T12:51:15

#2log_e(sqrtx+1)#

Note that #1/(x+sqrt(x))=1/(sqrtx(sqrtx+1))# and also that
#d/dx(sqrtx+1)=1/2(1/sqrtx)# so

#1/(sqrtx(sqrtx+1))=2(d/dx(sqrtx+1))/(sqrtx+1)# so

#intdx/(x+sqrt(x))=int1/(x+sqrt(x))dx=int2(d/dx(sqrtx+1))/(sqrtx+1)dx=2log_e(sqrtx+1)#

Answer 2 · 2016-11-18T12:53:45

The answer is #=2ln(1+sqrtx)+C#

The two writing represent the same integral.

Let #I=intdx/(x+sqrtx)#

Multiply numerator and denominator by #x-sqrtx#

#I=int((x-sqrtx)dx)/((x+sqrtx)(x-sqrtx))#

#=int((x-sqrtx)dx)/(x^2-x)#

Solving by substitution

Let #u=sqrtx# ;
#u^2=x#

#u^4 =x^2#

#du=dx/(2sqrtx)#

#dx=2udu#

#I=int(u^2-u)(2udu)/(u^4-u^2)#

#=2int(u^2(u-1)du)/(u^2(u^2-1))#

#=2int(cancel(u-1)du)/((u+1)cancel(u-1))#

#=2ln(u+1)#

#=2ln(1+sqrtx)+C#