The value of e 2 e 4 1 x ( e ( ( log e x ) 2 + 1 ) – 1 e ( ( log e x ) 2 + 1 ) – 1 + e ( ( 6 – log e x ) 2 + 1 ) – 1 ) d x is [2025]

Question

The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

Smart Achievers · Accepted Answer

(2)

Let $\log_{e}$ $x = t \Rightarrow \frac{1}{x} d x = d t$

$\Rightarrow I = \int_{2}^{4} \frac{e^{\frac{1}{t^{2} + 1}}}{e^{\frac{1}{t^{2} + 1}} + e^{\frac{1}{{(6 - t)}^{2} + 1}}} d t$ ... (i)

$\Rightarrow I = \int_{2}^{4} \frac{e^{\frac{1}{{(6 - t)}^{2} + 1}}}{e^{\frac{1}{{(6 - t)}^{2} + 1}} + e^{\frac{1}{t^{2 + 1}}}} d t$ $[∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (b + a - x) d x]$ ... (ii)

Adding (i) and (ii), we get

$2 I = \int_{2}^{4} d t = {[t]}_{2}^{4} = 4 - 2 = 2 \Rightarrow 2 I = 2 \Rightarrow I = 1$ .

Solution

Q.
The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

1 $\log_{e} 2$

2 1

3 $e^{2}$

4 2

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Solution

Q. The value of ∫e2e41x(e((logex)2+1)–1e((logex)2+1)–1+e((6–logex)2+1)–1)dx is [2025]

1 loge2

2 1

3 e2

4 2

Ans. (2) Let logex=t ⇒ 1xdx=dt ⇒ I=∫24e1t2+1e1t2+1+e1(6–t)2+1dt ... (i) ⇒ I=∫24e1(6–t)2+1e1(6–t)2+1+e1t2+1dt [∵ ∫abf(x)dx=∫abf(b+a–x)dx] ... (ii) Adding (i) and (ii), we get 2I=∫24dt=[t]24=4–2=2 ⇒ 2I=2 ⇒ I=1.

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Q.
The value of $\int_{e^{2}}^{e^{4}} \frac{1}{x} (\frac{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}}}{e^{({{(\log_{e} x)}^{2} + 1)}^{- 1}} + e^{({{(6 - \log_{e} x)}^{2} + 1)}^{- 1}}}) d x$ is [2025]

1 $\log_{e} 2$

3 $e^{2}$