Question #24569

1 Answer
mason m
Aug 7, 2016

#(2sqrt(x+a)(x-2a))/3+C#

Explanation:

We have:

#intx/sqrt(x+a)dx#

If we want to use substitution, a good bet would be to let #u=x+a#. This also would mean that #du=dx#. The only issue left is that we've accounted for both #dx# and #sqrt(x+a)# in the denominator, and still have the #x# in the numerator left over.

To deal with this, use the original substitution #u=x+a# to write #x# as #x=u-a#. Then, we have:

#intx/sqrt(x+a)dx=int(u-a)/sqrtudu#

At this point, split the integral through subtraction:

#int(u-a)/sqrtudu=intu/sqrtudu-inta/sqrtudu#

Rewrite using powers:

#intu/sqrtudu-inta/sqrtudu=intu^(1/2)du-aintu^(-1/2)du#

Integrate both using the power rule for integrals:

#intu^(1/2)du-aintu^(-1/2)du=u^(3/2)/(3/2)-a(u^(1/2)/(1/2))+C#

#=(2u^(3/2))/3-2au^(1/2)+C#

#=(2sqrtu(u-3a))/3+C#

Since #u=x+a#:

#=(2sqrt(x+a)(x+a-3a))/3+C#

#=(2sqrt(x+a)(x-2a))/3+C#

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