How do you integrate #int (x+1)sqrt(2-x)dx#?

Question

12012 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-10T21:36:43

#(-2(2-x)^(3/2)(x+3))/5+C#

#I=int(x+1)sqrt(2-x)dx#

Let #u=2-x#. This implies that #du=-dx#. Also note that #-u=x-2#, so #-u+3=x+1#. Then:

#I=int(-u+3)sqrtu(-du)=int(u-3)sqrtudu#

Expanding the square root as #u^(1/2)#:

#I=int(u(u^(1/2))-3u^(1/2))du=int(u^(3/2)-3u^(1/2))du#

Now using #intu^ndu=u^(n+1)/(n+1)+C#:

#I=u^(5/2)/(5/2)-3(u^(3/2)/(3/2))=2/5u^(5/2)-2u^(3/2)#

Factoring and making it look nice:

#I=u^(3/2)(2/5u-2)=(u^(3/2)(2u-10))/5=(2u^(3/2)(u-5))/5#

From #u=2-x#:

#I=(2(2-x)^(3/2)((2-x)-5))/5=(-2(2-x)^(3/2)(x+3))/5+C#