JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 4)

The integral $$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx$$, x > 0, is equal to : (where c is a constant of integration)
$${\log _e}\sqrt {{x^2} + 5x - 7} + c$$
$$4{\log _e}|{x^2} + 5x - 7| + c$$
$${1 \over 4}{\log _e}|{x^2} + 5x - 7| + c$$
$${\log _e}|{x^2} + 5x - 7| + c$$

Explanation

$$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}dx} $$

$$ = \int {{{8{x^3} + 5(4{x^2})} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$

$$ = \int {{{8{x^3} + 20{x^2}} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$

$$ = \int {{{8x + 20} \over {{x^2} + 5x - 7}}} $$

$$ = \int {{{4(2x + 5)} \over {{x^2} + 5x - 7}}} $$

Let $$\left\{ \matrix{ {x^2} + 5x - 7 = t \hfill \cr (2x + 5)dx = dt \hfill \cr} \right\}$$

$$ = \int {{{4dt} \over t}} $$

$$ = 4\ln \left| t \right| + C$$

$$ = 4\ln \left| {({x^2} + 5x - 7)} \right| + C$$

Here C is integral constant.

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