How do you integrate int (x+1)sqrt(2-x)dx∫(x+1)√2−xdx?

1 Answer
mason m
Dec 10, 2016

(-2(2-x)^(3/2)(x+3))/5+C−2(2−x)32(x+3)5+C

Explanation:

I=int(x+1)sqrt(2-x)dxI=∫(x+1)√2−xdx

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

I=int(-u+3)sqrtu(-du)=int(u-3)sqrtuduI=∫(−u+3)√u(−du)=∫(u−3)√udu

Expanding the square root as u^(1/2)u12:

I=int(u(u^(1/2))-3u^(1/2))du=int(u^(3/2)-3u^(1/2))duI=∫(u(u12)−3u12)du=∫(u32−3u12)du

Now using intu^ndu=u^(n+1)/(n+1)+C∫undu=un+1n+1+C:

I=u^(5/2)/(5/2)-3(u^(3/2)/(3/2))=2/5u^(5/2)-2u^(3/2)I=u5252−3(u3232)=25u52−2u32

Factoring and making it look nice:

I=u^(3/2)(2/5u-2)=(u^(3/2)(2u-10))/5=(2u^(3/2)(u-5))/5I=u32(25u−2)=u32(2u−10)5=2u32(u−5)5

From u=2-xu=2−x:

I=(2(2-x)^(3/2)((2-x)-5))/5=(-2(2-x)^(3/2)(x+3))/5+CI=2(2−x)32((2−x)−5)5=−2(2−x)32(x+3)5+C

Related questions
See all questions in Integration by Substitution
Impact of this question
12789 views around the world
You can reuse this answer
Creative Commons License