The integral \;{e³ _e (2x)}+5 e² _e (2x)}}{e^{4 _e x}+5 e³ _e x}-7 e² _e x}} dx, x > 0, is equal to (where, c is a constant of integration) [25 Feb 2021 Shift 2]

Correct Answer is : 4log⁡ex2+5x−7∣+c4 _e x²+5x-7|+c4logex2+5x−7∣+c

Correct Answer is : 4log⁡ex2+5x−7∣+c4 _e x²+5x-7|+c4logex2+5x−7∣+c

Test Index

Indefinite Integrals

Section: Mathematics

Question : 44 of 90

Marks: +1, -0

The integral

\int\;\frac{e^{3 \log_e (2x)}+5 e^{2 \log_e (2x)}}{e^{4 \log_e x}+5 e^{3 \log_e x}-7 e^{2 \log_e x}} dx

x > 0

c

Solution:

I=\int\;\frac{e^{3 \log_e(2x)}+5 e^{2 \log_e(2x)}}{e^{4 \log_e(x)}+5 e^{3 \log_e(x)}-7 e^{2 \log_e(x)}} d x

=\int\;\frac{e^{(\log_e(2x))^3}+5 e^{(\log_e(2x))^2}}{e^{\log_e(x^4)}+5 e^{(\log_e(x))^3}-7 e^{(\log_e(x))^2}} d x

[using property

a \log x = \log x^{a}

]

=\int\;\frac{8x^{3}+5(2x)^{2}}{x^{4}+5(x)^{3}-7x^{2}} d x \;\;

[using property

a^{\log_a x} = x]

\;\; =\int\;\frac{8x^{3}+20x^{2}}{x^{4}+5x^{3}-7x^{2}} d x = \int\;\frac{4x^{2}(2x+5)}{x^{2}(x^{2}+5x-7)} d x

\;\; =\int\;\frac{4(2x+5)}{x^{2}+5x+7} d x

Let

x^{2}+5x-7=t

, then

(2x+5) d x = d t

I=\int\;\frac{4 dt}{t}=4 \log_e t+c

Put

t=x^{2}+5x-7

I=4 \log_e |x^{2}+5x-7|+C

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