Check sibling questions

If ∫ (3e^x - 5e^(-x)) / (4e^x + 5e^(-x)) dx = ax + b log⁑ |4e^x + 5e^(-x) | + C, then

(A) a = (-1)/8, b = 7/8Β 

(B) a = 1/8, b = 7/8Β 

(C) a = (-1)/8, b = (-7)/8Β 

(D) a = 1/8, b = (-7)/8Β 

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Transcript

Question 3 If ∫1β–’γ€–(3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) )/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) 𝑑π‘₯=π‘Žπ‘₯+𝑏 log⁑〖|4𝑒^π‘₯+5𝑒^(βˆ’π‘₯) |γ€—+𝐢〗, then π‘Ž=(βˆ’1)/8, 𝑏=7/8 (B) π‘Ž=1/8, 𝑏=7/8 (C) π‘Ž=(βˆ’1)/8, 𝑏=(βˆ’7)/8 (D) π‘Ž=1/8, 𝑏=(βˆ’7)/8 To solve this, We write (πŸ‘π’†^𝒙 βˆ’ πŸ“π’†^(βˆ’π’™) )/(πŸ’π’†^𝒙 + πŸ“π’†^(βˆ’π’™) ) as (3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) )/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) = a + b Γ— (𝐝(πŸ’π’†^𝒙 + πŸ“π’†^(βˆ’π’™) )/𝒅𝒙)/(πŸ’π’†^𝒙 + πŸ“π’†^(βˆ’π’™) ) (3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) )/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) = a + b Γ— (4𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯))/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) (3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) )/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) = (π‘Ž Γ— (4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) + 𝑏 Γ— (4𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯)))/(4𝑒^π‘₯ + 5𝑒^(βˆ’π‘₯) ) 3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) = π‘Ž Γ— (4𝑒^π‘₯+ 5𝑒^(βˆ’π‘₯) ) + 𝑏 Γ— (4𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯)) 3𝑒^π‘₯ βˆ’ 5𝑒^(βˆ’π‘₯) = 4π‘Ž 𝑒^π‘₯+5π‘Ž 𝑒^(βˆ’π‘₯)+4𝑏 𝑒^π‘₯βˆ’5𝑏 𝑒^(βˆ’π‘₯) πŸ‘π’†^𝒙 βˆ’ πŸ“π’†^(βˆ’π’™) = 𝒆^𝒙 (πŸ’π’‚+πŸ’π’ƒ)+𝒆^(βˆ’π’™) (πŸ“π’‚βˆ’πŸ“π’ƒ) πŸ‘ = πŸ’π’‚+πŸ’π’ƒ 4π‘Ž+4𝑏=3 𝒂+𝒃=3/4 βˆ’πŸ“ = πŸ“π’‚βˆ’πŸ“π’ƒπ’ƒ 5π‘Žβˆ’5𝑏=βˆ’5 π’‚βˆ’π’ƒ=βˆ’1 Thus, our equations are 𝒂+𝒃=3/4 …(1) π’‚βˆ’π’ƒ=βˆ’1 …(2) Adding (1) and (2) π‘Ž+𝑏+π‘Žβˆ’b=3/4βˆ’1 2a=(βˆ’1)/4 𝐚=(βˆ’πŸ)/πŸ– Putting value of a in (1) π‘Ž+𝑏=3/4 (βˆ’πŸ)/πŸ–+𝒃=πŸ‘/πŸ’ 𝑏=3/4+1/8 𝑏=6/8+1/8 𝒃=7/8 Thus, 𝐚=(βˆ’πŸ)/πŸ– and 𝒃=7/8 So, the correct answer is (a)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.