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3.4 · Q1

Q.Evaluate the following definite integral: ∫ee21xlog⁡x dx\int_e^{e^2} \frac{1}{x\log x}\,dx

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

Substitute t=log⁡xt=\log x; the definite integral becomes ∫12dtt=log⁡2\int_1^2\frac{dt}{t}=\log 2.

Substitution: if t=log⁡xt=\log x then dt=dxxdt=\frac{dx}{x}, and limits transform with xx.

  1. Given: ∫ee2dxxlog⁡x\displaystyle\int_e^{e^2}\frac{dx}{x\log x}.
  2. Put t=log⁡x⇒dt=dxxt=\log x\Rightarrow dt=\dfrac{dx}{x}. Limits: x=e⇒t=1x=e\Rightarrow t=1; x=e2⇒t=2x=e^2\Rightarrow t=2.
  3. =∫12dtt=[log⁡t]12=log⁡2−log⁡1=log⁡2=\displaystyle\int_1^2\frac{dt}{t}=\big[\log t\big]_1^2=\log 2-\log 1=\log 2.
  4. Value: log⁡2≈0.693\log 2\approx 0.693.
✓Final answer

∫ee2dxxlog⁡x=log⁡2≈0.693\displaystyle\int_e^{e^2}\frac{dx}{x\log x}=\log 2\approx 0.693.

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