Can someone please help me to solve
$$\int_{2}^{\infty} \frac{x^2-x}{6^x} \mathrm dx ?$$
I don't know how to integrate it as there is no exponential involved but the constant $6$.
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2There is exactly one real number such that, for all real numbers $x$, the idenity $e^{\alpha x}=\frac1{6^x}$ holds. The question for you is: which number is $\alpha$? – Mar 06 '17 at 04:29
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1Have you considered integration by parts plus the fact that for some constant $b > 0$, $\frac{d}{dx} b^x = b^x \cdot \ln(b)$? – AlkaKadri Mar 06 '17 at 04:29
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@G.Sassatelli I didn't change $1/6^{x}$ to $6^{-x}$? – Matthew Cassell Mar 06 '17 at 04:34
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@Mattos, he's is essentially trying to get OP to figure out how to differentiate $6^x$ knowing that $\frac{d}{dx} e^x = e^x$ and the chain rule. More generally, notice that $\frac{d}{dx} b^x = \frac{d}{dx} e^{\ln(b)x} = e^{\ln(b)x} \cdot \ln(b) = b^x \cdot \ln(b)$ – AlkaKadri Mar 06 '17 at 04:46
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@AlkaKadri Mate, G.Sassatelli and I were talking about an edit that was made to the question that he mistakenly attributed to me. He then removed his comment about the edit. Our exchange had nothing to do with understanding G.Sassatellis comment to the OP which is still visible. – Matthew Cassell Mar 06 '17 at 04:57
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@Mattos ah my bad bro, a misunderstanding on my part. – AlkaKadri Mar 06 '17 at 05:07
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@AlkaKadri All good mate. – Matthew Cassell Mar 06 '17 at 05:14
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Let us continue this discussion in chat. – AlkaKadri Mar 06 '17 at 05:33
1 Answers
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Expanding on the comments. There is an exponential function.
Recall that for $a>0$,
$$a^x = (e^{\ln a})^x = e^{x\ln (a) },$$
so the integral can be rewritten as
$$\int_{2}^{\infty} (x^2-x)e^{-x\ln(6)} \mathrm dx.$$
Got it from there?
zahbaz
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