How do you evaluate the integral #int (1-sqrtx)/(1+sqrtx)#?

1 Answer
Noah G
Feb 23, 2017

#-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C#

Explanation:

Use partial fractions to get rid of the radical in the numerator.

#A/1 + B/(1 + sqrt(x)) = (1 - sqrt(x))/(1 + sqrt(x))#

#A(1 + sqrt(x)) + B = 1 - sqrt(x)#

#A + Asqrt(x) + B = 1 - sqrt(x)#

This means that #{(A = -1), (A + B = 1):}#

Solving, we get that #A = -1# and #B = 2#. The integral becomes

#int -1 + 2/(1 + sqrt(x))dx#

#int -1dx + int 2/(1 + sqrt(x))dx#

#int -1dx + 2int 1/(1 + sqrt(x))dx#

Let #u = 1 + sqrt(x)#. Then #du = 1/(2sqrt(x))dx# and #dx = 2sqrt(x)du#. We can also conclude that #2sqrt(x) = 2(u - 1)#, because #sqrt(x) = u -1# and we have two of those.

#int -1dx + 2int 1/u * 2(u - 1)du#

#int -1dx + 4int (u - 1)/u du#

#int -1dx + 4int u/udu - 4int 1/udu#

#int -1dx + 4int 1du- 4int 1/u du#

#-x + 4u - 4ln|u| + C#

#-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C#

Hopefully this helps!

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