Evaluate- - e a log x - e x log a d x

The correct option is B x ^ a + 1 a + 1 + a ^ x log a + c ∫ e^alog x+e^xlog adx =∫ e^log x^a+e^log a^xdx =∫ x^a+a^xdx =x^a+1a+1+ ...

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Special Integrals - 1

Evaluate: ∫...

Question

Evaluate: $\int (e^{a log x} + e^{x log a}) d x$

x^{a + 1} + \frac{a^{x}}{log a} + c

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\frac{x^{a + 1}}{a + 1} + a^{x} log a + c

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\frac{x^{a + 1}}{a + 1} + \frac{a^{x}}{log a} + c

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\int a^{x} d x = \frac{a^{x}}{log a} + c

\frac{1}{1 + x^{(b - a)} + x^{(c - a)}} + \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} + \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ?

$\frac{1}{1 + x^{(b - a)} + x^{(c - a)}} . \frac{1}{1 + x^{(a - b)} + x^{(c - b)}} . \frac{1}{1 + x^{(b - c)} + x^{(a - c)}} = ?$

Prove that:

(i) $\frac{1}{1 + x^{a - b}} + \frac{1}{1 + x^{b - a}} = 1$

(ii) $\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{b - c} + x^{a - c}} = 1$

\frac{1}{1 + x^{b - a} + x^{c - a}} + \frac{1}{1 + x^{a - b} + x^{c - b}} + \frac{1}{1 + x^{a - c} + x^{b - c}}

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