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$$g(x) = \int_{2x}^{6x} \frac{u+2}{u-4}du $$

For finding the $ g'(x)$, would I require to find first the derivative of $\frac{u+2}{u-4}$

then Replace the $u$ with 6x and 2x and add them ? (the 2x would have to flip so the whole term is negative)

If the previous statement is true would the final showdown be the following: $$ \frac{6}{(2x-4)^2} - \frac{6}{(6x-4)^2}$$

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I think you do not need to differentiate the integrand. Hint: if $f(x) = \int_0^x \frac{u+2}{u-4}du$, then $f'(x) = \frac{x+2}{x-4}$, so try to express $g$ in terms of $f$. – Lior B-S Jan 28 '13 at 7:06
Are you saying: $$6\frac{6x+2}{6x-4}+2\frac{2x+2}{2x-4} $$?? – Reza M. Jan 28 '13 at 7:09
Almost right, but minus, not plus. – André Nicolas Jan 28 '13 at 7:12
up vote 3 down vote accepted

Let $f(u)=\frac{u+2}{u-4}$, and let $F(u)$ be the antiderivative of $f(u)$. Then $$ g'(x)=\frac{d}{dx}\int_{2x}^{6x}f(u)du=\frac{d}{dx}\left(F(u)\bigg\vert_{2x}^{6x}\right)=\frac{d}{dx}[F(6x)-F(2x)]=6F'(6x)-2F'(2x) $$ But $F'(u)=f(u)$. So the above evaluates to $$ 6f(6x)-2f(2x)=6\frac{6x+2}{6x-4}-2\frac{2x+2}{2x-4}=\cdots. $$ In general, $$ \frac{d}{dx}\int_{a(x)}^{b(x)}f(u)du=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x). $$

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Oh! I understand, the dx(derivative) cancels out the integral(Anti-Derivative) So it stays the same. Then all that is required is to plug and play the correct areas. – Reza M. Jan 28 '13 at 7:15
@RezaM. Taking the derivative more or less "undoes" the anti-derivative, but in general you have to be careful about the bounds of integration. – Ben West Jan 28 '13 at 7:17

According to Leibniz integral rule $$g'(x)=f(6x)\times 6-f(2x)\times 2$$ wherein $f(t)=\large \frac{t+2}{t-4}$.

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+1 I just made your fraction a teeny bit bigger! :-) – amWhy Jan 28 '13 at 16:10

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