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Question: \(\int_{}^{}\frac{(a^{x} + b^{x})^{2}}{a^{x}b^{x}}\) dx is equal to –...

∫(ax+bx)2axbx\int_{}^{}\frac{(a^{x} + b^{x})^{2}}{a^{x}b^{x}} dx is equal to –

A

(b/a)xlog⁡e(a/b)\frac{(b/a)^{x}}{\log_{e}(a/b)}+ (a/b)xlog⁡e(b/a)\frac{(a/b)^{x}}{\log_{e}(b/a)} + x + c

B

(a/b)xlog⁡e(b/a)\frac{(a/b)^{x}}{\log_{e}(b/a)} + (b/a)xlog⁡e(a/b)\frac{(b/a)^{x}}{\log_{e}(a/b)} + x + c

C

(a/b)xlog⁡e(b/a)\frac{(a/b)^{x}}{\log_{e}(b/a)} + (b/a)xlog⁡e(a/b)\frac{(b/a)^{x}}{\log_{e}(a/b)} + 2x + c

D

(a/b)xlog⁡e(a/b)\frac{(a/b)^{x}}{\log_{e}(a/b)} + (b/a)xlog⁡e(b/a)\frac{(b/a)^{x}}{\log_{e}(b/a)} + 2x + c

Answer

(a/b)xlog⁡e(a/b)\frac{(a/b)^{x}}{\log_{e}(a/b)} + (b/a)xlog⁡e(b/a)\frac{(b/a)^{x}}{\log_{e}(b/a)} + 2x + c

Explanation

Solution

∫(ax+bx)2axbx\int \frac { \left( \mathrm { a } ^ { \mathrm { x } } + \mathrm { b } ^ { \mathrm { x } } \right) ^ { 2 } } { \mathrm { a } ^ { \mathrm { x } } \mathrm { b } ^ { \mathrm { x } } } dx

= ∫a2x+b2x+2axbxaxbx\int \frac { a ^ { 2 x } + b ^ { 2 x } + 2 a ^ { x } b ^ { x } } { a ^ { x } b ^ { x } } dx

= dx

= ∫{(ab)x+(ba)x+2}\int \left\{ \left( \frac { \mathrm { a } } { \mathrm { b } } \right) ^ { \mathrm { x } } + \left( \frac { \mathrm { b } } { \mathrm { a } } \right) ^ { \mathrm { x } } + 2 \right\} dx

= (a/b)xlog⁡e(a/b)\frac { ( a / b ) ^ { x } } { \log _ { e } ( a / b ) }+ (b/a)xlog⁡e(b/a)\frac{(b/a)^{x}}{\log_{e}(b/a)} + 2x + c.

Hence (4) is the correct answer.