Mathematics

$$\int(e^{a \log x}+ e^{x \log a})dx$$


SOLUTION
$$\int (e^{a \log x}+e^{x \log a})dx$$
Assume $$\log$$ on base $$e$$
$$a \log x = ln x^{a}$$
$$x \log a= ln a^{x}$$
$$=\int (e^{lnx^{a}}+e^{lna^{x}})dx$$
we know that $$e^{ln(f(x))}=f(x)$$
$$=\int (x^{a}+a^{x})dx$$
$$=\dfrac{x^{a+1}}{a+1}+\dfrac{a^{x}}{ln \ a}+c$$
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Subjective Medium Published on 17th 09, 2020
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