∫ax+bx2axbxdx=∫a2x+b2x+2axbxaxbxdx=∫axbx+bxax+2dx=∫abxdx+∫baxdx+2∫1dx=(a/b)xlog⁡(a/b)+(b/a)xlog⁡(b/a)+2x+c=(a/b)xlog⁡a−log⁡b+(b/a)xlog⁡b−log⁡a+2x+c=(a/b)x−(b/a)xlog⁡a−log⁡b+2x+c

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$\int \frac{{(a^{x} + b^{x})}^{2}}{a^{x} b^{x}} d x =$

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$\frac{(a / b)^{x} + (b / a)^{x}}{\log a + \log b} - 2 x + c$

$\frac{(a / b)^{x} - (b / a)^{x}}{\log a - \log b} - 2 \dot{x} + c$

$\frac{(a / b)^{x} - (b / a)^{x}}{\log a - \log b} + 2 x + c$

$\frac{(a / b)^{x} + (b / a)^{x}}{\log a + \log b} + 2 x + c$

answer is C.

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Detailed Solution

$\begin{aligned} \int \frac{{(a^{x} + b^{x})}^{2}}{a^{x} b^{x}} d x = \int (\frac{a^{2 x} + b^{2 x} + 2 a^{x} b^{x}}{a^{x} b^{x}}) d x = \int (\frac{a^{x}}{b^{x}} + \frac{b^{x}}{a^{x}} + 2) d x = \int {(\frac{a}{b})}^{x} d x + \int {(\frac{b}{a})}^{x} d x + 2 \int 1 d x \\ = \frac{(a / b)^{x}}{\log (a / b)} + \frac{(b / a)^{x}}{\log (b / a)} + 2 x + c = \frac{(a / b)^{x}}{\log a - \log b} + \frac{(b / a)^{x}}{\log b - \log a} + 2 x + c = \frac{(a / b)^{x} - (b / a)^{x}}{\log a - \log b} + 2 x + c \end{aligned}$