The integral int frac{e^{3 log_{e} 2x} + 5e^{ | vedclass

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Question

(B) Given integral: $I = \int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$Using the pro

Vedclass · Accepted Answer

$4 \log_{e} |x^{2} + 5x - 7| + c$

Vedclass · Answer

(B) Given integral: $I = \int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$Using the property $e^{\log_{e} f(x)} = f(x)$,we have:$I = \int \frac{(2x)^{3} + 5(2x)^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$$I = \int \frac{8x^{3} + 20x^{2}}{x^{4} + 5x^{3} - 7x^{2}} dx$Factor out $4x^{2}$ from the numerator and $x^{2}$ from the denominator:$I = \int \frac{4x^{2}(2x + 5)}{x^{2}(x^{2} + 5x - 7)} dx$$I = 4 \int \frac{2x + 5}{x^{2} + 5x - 7} dx$Let $u = x^{2} + 5x - 7$,then $du = (2x + 5) dx$.$I = 4 \int \frac{1}{u} du = 4 \log_{e} |u| + c$$I = 4 \log_{e} |x^{2} + 5x - 7| + c$

The integral $\int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$,where $x > 0$,is equal to ....... (where $c$ is a constant of integration).

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