I can't seem to figure out how to integrate 2x ---- dx ? x-4 I tried U substitution for both 2x and x-4 but to no avail
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Originally Posted by nottheface I can't seem to figure out how to integrate 2x ---- dx ? x-4 I tried U substitution for both 2x and x-4 but to no avail Hint: $\displaystyle \dfrac{2x}{x-4}=\dfrac{2x-8+8}{x-4}=\dfrac{2(x-4)+8}{x-4}=2+\dfrac{8}{x-4}$
Another way to do the same thing: substitute u= x- 4 as you did. Then x= u+ 4 so 2x= 2u+ 8. du= dx and the integral becomes $\displaystyle \int \frac{2u+ 8}{u}du= \int \left(2+ \frac{8}{u}\right) du$
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