Question

integrate (e^3a log x + e^3x log a) dx.​

Answer 1

Given that

$\displaystyle\sf y = \int e^{3a \cdot log \ x} + e^{3x \cdot log \ a} dx$

Solution:-

we know that $\sf a^{mn} = (a^m)^n$

$\displaystyle\sf y = \int (e^{3a \cdot log \ x} + e^{3x \cdot log \ a}) dx$

$\displaystyle\sf = \int e^{(log \ x)3a} dx + \int e^{(log \ a)3x} dx$

$\displaystyle\sf = \int x^{3a} dx + \int a^{3x} dx$

$\boxed{\sf = \dfrac{x^{3a} +1}{3a + 1} + \dfrac{1}{3} a^{3x} \ log \ a + C}$

know more!

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

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