$\int (e^{a log x} + e^{x log a}) d x$ is equal to.

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

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

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x^{a + 1} + a^{x} + c

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

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Solution

The correct option is A $\frac{x^{a + 1}}{a + 1} + \frac{a^{x}}{log a} + c$
$\int (e^{a log x} + e^{x log a}) d x$
$= \int [e^{log x^{a}} + e^{log a^{x}}] d x$
$= \int [x^{a} + a^{x}] d x$

$= \int x^{a} d x + \int a^{x} d x$
$= \frac{x^{a + 1}}{a + 1} + \frac{a^{x}}{log a} + c$

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Similar questions

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

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

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

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