∫ 1 4 x + x + 1 x d x - ∫ 1 2 log 2 d x =

Question

∫ 1 4 x + x + 1 x d x - ∫ 1 2 log 2 d x =, 79 6, 643 6, 321 5, 64

MARKS App · Accepted Answer

Let I = ∫ 1 4 x + x + 1 x d x - ∫ 1 2 log 2 d x ⇒ I = x 2 2 + 2 3 x 3 2 + log x 1 4 - x 1 2 log 2 ⇒ I = 16 2 + 2 × 8 3 + log 4 - 1 2 2 + 2 3 + log 1 - 2 log 2 - 1 ⇒ I = 15 2 + 14 3 + 2 log 2 - 2 log 2 - 1 ⇒ I = 79 6

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