Integrate $\int \frac{x-6}{\sqrt{x^2-12x+9}} dx$ using u substitution

Question

Section 5.2

Can somebody verify this solution for me? Thanks!

Integrate $\int \frac{x-6}{\sqrt{x^2-12x+9}} dx$ using u substitution

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Let $u=x^2-12x+9$. Then $\frac{du}{dx}=2x-12$ and so $\frac{du}{2x-12}=dx$.

Thus we have:

$\int \frac{x-6}{\sqrt{x^2-12x+9}} dx$

$= \int \frac{x-6}{\sqrt{u}}\frac{du}{2x-12}$

$= \int \frac{x-6}{\sqrt{u}}\frac{du}{2(x-6)}$

$=\frac{1}{2} \int \frac{du}{\sqrt{u}}$

$=\frac{1}{2} \int u^{\frac{-1}{2}}du$

$=\frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C$

$=u^{\frac{1}{2}}+C$

$=(x^2-12x+9)^{\frac{1}{2}}+C$

Answer 1

Your solution is correct. Could have reduced the number of steps by directly substituting $\ t^2=x^2-12x+9$ and $(x-6)dx=tdt$ to obtain $I=\int \frac{tdt}{t}=\int dt=t+C$, but the solution would have essentially remained the same.