How do you integrate #int_0^1 1/(1+(x)^(1/2)) dx# ?

2 Answers
Feb 14, 2018

Write #x^(1/2)# as #sqrtx#:

#int_0^1 1/(1+sqrtx) dx#

To do the indefinite integral, I used a tip from WolframAlpha :

Let #u = sqrtx# then #du = 1/(2sqrtx)dx#

We need to solve for #dx#:

#dx = 2sqrtxdu#

But #sqrtx = u#:

#dx = 2udu#

Perform the substitutions:

#int 1/(1+(x)^(1/2)) dx = 2 int u/(u+1) du#

Add 0 to the numerator in the form if #+1 - 1#

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

Separate into two fractions:

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

The first fraction becomes 1:

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

Both terms are trival to integrate:

#int 1/(1+(x)^(1/2)) dx = 2 (u-ln(u+1))+C#

Reverse the substitution:

#int 1/(1+(x)^(1/2)) dx = 2 (sqrtx-ln(sqrtx+1))+C#

To integrate from 0 to 1 we evaluate the right side at #x = 1# and subtract the right side evaluated at #x = 0#:

#int_0^1 1/(1+(x)^(1/2)) dx = {2(sqrt1-ln(sqrt1+1))+C}-{2(sqrt0-ln(sqrt0+1))+C}#

The constants of integration sum to 0 and #sqrt0 - ln(1) = 0#, therefore, the answer is:

#int_0^1 1/(1+(x)^(1/2)) dx = 2-2ln(2)#

Feb 14, 2018

#f(x) = 1/(1+sqrtx)#

let #sqrtx=t#
#x=t^2#
differentiating both the sides,
#dx = 2t.dt#

#intf(x) = int1/(1+sqrtx) dx#
#= int1/(1+t) 2t. dt = int(2t)/(1+t) dt#
applying simple polynomial division,
#int (2-(2)/(1+t)) dt#
#int 2.dt-int(2)/(1+t) dt#
#2intdt-2int1/(1+t) dt#
#2t-2ln(1+t) +c#

replacing, #t=sqrtx#
#2sqrtx-2ln(1+sqrtx) +c#

now, applying the limits,
#2sqrt1-2ln(1+sqrt1) - (2sqrt0-2ln(1+sqrt0))#
#2-2ln(1+1) - (-2ln(1))# as #color(purple)(sqrt1=1; sqrt0=0;ln1=0#
#2-2ln2 #