$$\displaystyle \int6^{-2x}dx$$
I got $-\dfrac{6^{-2x}}{2\ln6}$ but I'm not confident that its correct.
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1Our function is $e^{-x(2\ln 6)}$. So you are right, apart from the missing $+C$. – André Nicolas Apr 24 '14 at 01:05
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1differentiate your result, if you get the same function you are integrating, then your result is correct. – yaa09d Apr 24 '14 at 01:59
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Let $y=6^{-2x}$, then \begin{align} \ln y=\ln\left(6^{-2x}\right)=-2x\ln6\qquad\rightarrow\qquad y=e^{-(2\ln6)x} \end{align} Hence \begin{align} \int6^{-2x}\,dx&=\int e^{-(2\ln6)x}\,dx \end{align} Remember \begin{align} \int e^{ax}\,dx=\frac{e^{ax}}{a}+C \end{align} then \begin{align} \int6^{-2x}\,dx&=\int e^{-(2\ln6)x}\,dx\\ &=\frac{e^{-(2\ln6)x}}{-(2\ln6)}+C\\ &=-\frac{e^{-(2\ln6)x}}{2\ln6}+C\\ &=-\frac{6^{-2x}}{2\ln6}+C\\ \end{align}
Anastasiya-Romanova 秀
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Our integral is: $$\int 6^{-2x} \ dx$$ Rewrite the integral: $$\int \left(6^{-2}\right)^x \ dx$$ $$=\int \left(\frac{1}{36}\right)^x \ dx$$ Remember that: $$\int a^x \ dx=\frac{a^x}{\ln a}+C$$ In this case our $a$ is $\frac{1}{36}$ $$\int \left(\frac 1{36}\right)^x \ dx=\frac{\left(\frac{1}{36}\right)^x}{\ln \left(\frac{1}{36}\right)}+C$$ $$=\frac{6^{-2x}}{\ln\left(6^{-2}\right)}+C$$ $$=-\frac{6^{-2x}}{2 \ln 6}+C$$ Hope I helped!
Anonymous Computer
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what was the purpose of rewriting $6^{-2}$ as $\frac{1}{36}$ if you just substituted back in as soon as you integrated? looks nicer i guess – MT_ Apr 24 '14 at 01:38
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1@MichaelT I don't have the ability to look ahead and see how my answer would turn out (note: not sarcasm). Both ways are fine I think – Anonymous Computer Apr 24 '14 at 01:40