U substitution indefinite integral

**Chewy** · October 6th 2012, 05:21 AM

I have been trying to solve the indefinite integral (x-3)*sqrt(x-1) dx. I have tried u substitution by using u=(x-1)^1/2 or using u=x-3. There has to be a trick to this problem that I'm not seeing and it's driving me nuts!

This is a u substitution problem right? Any help is greatly appreciated!

**skeeter** · October 6th 2012, 05:39 AM

Originally Posted by Chewy

I have been trying to solve the indefinite integral (x-3)*sqrt(x-1) dx. I have tried u substitution by using u=(x-1)^1/2 or using u=x-3. There has to be a trick to this problem that I'm not seeing ...

$u = x-1$

$du = dx$

$x = u+1$

$\int (x-3)\sqrt{x-1} \, dx$

substitute ...

$\int [(u+1)-3] \sqrt{u} \, du$

proceed ...

**Soroban** · October 6th 2012, 12:48 PM

Hello, Chewy!

$\int (x-3)\sqrt{x-1}\,dx$

If the expression under the radical is linear,
. . we can let $u$ = the entire radical.

Let $u \:=\:\sqrt{x-1}\quad\Rightarrow\quad x-1 \:=\:u^2 \quad\Rightarrow\quad dx \:=\:2u\,du$

Substitute: . $\int (u^2-2)\cdot u \cdot 2u\,du \;=\;2\int(u^4 - 2u^2)\,du$

. . . . . . . . . . $=\;2\left(\tfrac{1}{5}u^5 - \tfrac{2}{3}u^3\right) + C \;=\; \tfrac{2}{15}u^3\left(3u^2-10)+C$

Back-substitute: . $\tfrac{2}{15}\left(\sqrt{x-1}\right)^3\big[3(x-1) - 10\big] + C$

. . . . . . . . . . . $=\;\tfrac{2}{15}(x-1)^{\frac{3}{2}}(3x-13) + C$

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