Integrate x*(x-1)^(1/2) which is translated to Int

[quazzi]

The starting example shows u= x-1 du=dx that I get bur then it converts the integral to u^(1/2)*(u+1)du ??

the end result should be in the for of (2U^(5/2))/5 + (2U^(3/2))/3 + C

thanks

 

[galactus]

Try this:

Let \(\displaystyle \L\\u=\sqrt{x-1}; \;\ u^{2}+1=x; \;\ 2udu=dx\)

Sub in appropriately:

\(\displaystyle \L\\\int{2u(u^{2}+1)udu=\int{2u^{4}+2u^{2}}du\)

 

[soroban]

Hello, quazzi!

I'll do it their way . . .

\(\displaystyle \L\int x\sqrt{x\,-\,1}\,dx\)

They used: \(\displaystyle \,u\:=\:x\,-\,1\;\;\Rightarrow\;\;x\:=\:u\,+\,1\;\;\Rightarrow\;\;dx\:=\:du\)
. . And: \(\displaystyle \x\,-\,1)^{\frac{1}{2}}\:=\:u^{\frac{1}{2}}\)


Substitute: \(\displaystyle \L\:\int \underbrace{x}_{\downarrow}\,\underbrace{(x\,-\,1)^{\frac{1}{2}}}_{\downarrow}\,\underbrace{dx}_{\swarrow}\)
. . . . . . \(\displaystyle \L\:\int(u\,+\,1)\cdot u^{\frac{1}{2}}\cdot du\)

We have: \(\displaystyle \L\:\int\left(u^{\frac{3}{2}}\,+\,u^{\frac{1}{2}}\right)\,du\)

Then: \(\displaystyle \L\:\frac{2}{5}u^{\frac{5}{2}}\,+\,\frac{2}{3}u^{\frac{3}{2}}\,+\,C\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We're not finished . . . We must back-substitute.
. . But I always simplify first.

Factor: \(\displaystyle \L\:\frac{u^{\frac{3}{2}}}{15}\left(6u\,+\,10\right)\,+\,C\)


Replace \(\displaystyle u\) with \(\displaystyle x\,-\,1\)

. . \(\displaystyle \L\frac{(x\,-\,1)^{\frac{3}{2}}}{15}\left[6(x\,-\,1)\,+\,10\right]\,+\,C \;=\;\frac{(x\,-\,1)^{\frac{3}{2}}}{15}(6x\,+\,4)\,+\,C\)

. . \(\displaystyle \L=\;\frac{2}{15}(x\,-\,1)^{\frac{3}{2}}(3x\,+\,2)\,+\,C\)


This is probably the answer at the back of the book.
. . They just love to simplify beyond all recognition.