_{ 4}^{ 5} {e²x}+e^x}{e²x}-5e^x+6} dx = [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : log⁡(12827) ≤ft( 128/27 )log(27128)

Correct Answer is : log⁡(12827) ≤ft( 128/27 )log(27128)

Test Index

Definite Integration

Section: Mathematics

Question : 52 of 54

Marks: +1, -0

\int\limits_{\log 4}^{\log 5} \frac{e^{2x}+e^x}{e^{2x}-5e^x+6} dx =

Solution:

\text{ Let } I = \int\limits_{\log 4}^{\log 5} \frac{e^{2x}+e^x}{e^{2x}-5e^x+6} dx

= \int\limits_{\log 4}^{\log 5} \frac{(e^x)^2 + e^x}{(e^x-3)(e^x-2)} dx

= \int\limits_{\log 4}^{\log 5} \frac{e^x(e^x+1)}{(e^x-3)(e^x-2)} dx

Let

e^x = t

\Rightarrow e^x dx = dt

When

x \rightarrow \log 4, t \rightarrow 4

x \rightarrow \log 5, t \rightarrow 5

I = \int\limits_{4}^{5} \frac{t+1}{(t-3)(t-2)} dt

= \int\limits_{4}^{5} \frac{4}{t-3} - \frac{3}{t-2}

[ ∵ by partial fraction]

= \left[ 4 \log |t-3| - 3 \log |t-2| \right]_4^5

= \left[ \log \left\| \frac{(t-3)^4}{(t-2)^3} \right\|. \right]_4^5

= \log \frac{2^4}{3^3} - \log \frac{1}{2^3} = \log \frac{2^4}{3^3} \times 2^3

= \log \left( \frac{2^7}{3^3} \right) = \log \left( \frac{128}{27} \right)

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