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How do you integrate

+1 vote

 

(8t+4)^2/(t^2)?

asked Jan 17, 2013 in CALCULUS by angel12 Scholar

2 Answers

+1 vote

∫ (8t + 4)^2 / (t^2) dt

[ Note : (a + b)^2 = a^2 + 2ab + b^2 ]

∫ [( (8t)^2 + 2(8t)(2) + (4)^2 ) / t^2] dt

Simplify

∫ [ (64(t^2) + 32t + 16) / t^2 ]dt

[ Note : (a+b)/c = a/c + b/c ]

∫ [ (64(t^2)/(t^2) + 32t/(t^2) + 16/(t^2)]dt

Simplify

∫ [ 64 + 32/t + 16/(t^2)]dt

Simplify

∫ 64 dt + ∫ (32/t)dt + ∫16/(t^2)dt

64∫ dt + 32∫(1/t) dt + 16∫(1/t^2) dt

[ Formulas : ∫ 1dx = x + c  ,  ∫(1/x) dx = log x + c  ,  ∫(1/t^2) dt = (-1 / x) + c ]

64 t + 32 log t + 16(-1/ t) + c                  where c is constant

64 t + 32 log t - 16/t + c

answered Jan 17, 2013 by richardson Scholar

∫ (8t + 4)^2 / (t^2) dt = 64t - 16/t + 64logt + C

0 votes

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image is the solution.

answered Jul 3, 2014 by joly Scholar

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