∫ 4 e x - 25 2 e x - 5 d x = A x + B log ⁡ | 2 e x - 5 | + c then

Question

∫ 4 e x - 25 2 e x - 5 d x = A x + B log ⁡ | 2 e x - 5 | + c then, A = 5 , B = 3, A = 5 , B = - 3, A = - 5 , B = 3, A = - 5 , B = - 3

MARKS App · Accepted Answer

Let I = 4 e x - 25 2 e x - 5 d x ⇒ I = ∫ 10 e x - 25 - 6 e x 2 e x - 5 d x ⇒ I = ∫ 5 2 e x - 5 - 6 e x 2 e x - 5 d x ⇒ I = ∫ 5 - 6 e x 2 e x - 5 d x ⇒ I = 5 x - 3 log ⁡ 2 e x - 5 + C ∴ A = 5 a n d B = - 3

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