Hello, mattflint50!
i was given: \(\displaystyle \L\: \int x\sqrt{x\,+\,3}\,dx\)
I have tried numerous times to do U substitution, but I do not know what to let U =
What did you try? . . . What did you get? . . . Where did you get stuck?
If you understand U-substitution, most choices will work.
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I prefer to eliminate the radical if possible . . .
Let \(\displaystyle u\,= \sqrt{x\,+\,3}\;\;\Rightarrow\;\;x\:=\:u^2\,-\,3\;\;\Rightarrow\;\;dx\,=\,2u\,du\)
We have: \(\displaystyle \L\;\;\;\;\;\;\int x\,\underbrace{\sqrt{x\,+\,3}}\,dx\)
. . . . . . . . . . . . . . . . . . \(\displaystyle \downarrow\;\;\;\;\:\downarrow\;\;\;\;\;\downarrow\)
Substitute: \(\displaystyle \L\:\:\int (u^2\,-\,3)\cdot u\cdot(2u\,du)\)
And we have: \(\displaystyle \L\:2\int(u^4\,-\,3u^2)\,du\;\) to integrate.
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Unco's suggestion will also work . . .
Let \(\displaystyle u\,=\,x\,+\,3\;\;\Rightarrow\;\;x\:=\:u\,-\,3\;\;\Rightarrow\;\;dx\,=\,du\)
We have: \(\displaystyle \L\;\;\;\int x\cdot\sqrt{x\,+\,3}\cdot dx\)
. . . . . . . . . . . . . . . \(\displaystyle \downarrow\;\;\;\;\;\downarrow\;\;\;\downarrow\)
Substitute: \(\displaystyle \L\:\int(u\,-\,3)\cdot u^{\frac{1}{2}}\cdot du\)
And we have: \(\displaystyle \L\:\int\left(u^{\frac{3}{2}}\,-\,3u^{\frac{1}{2}\right)\,du\;\) to integrate.
So exactly
where is your difficulty?
