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PSYCHO15rus [73]

3 years ago

12

Evaluate the integral ∫_0^1{36dx/((2x+1)^3)}

1 answer:

german3 years ago

3 0

$36 \int\limits^1_0 { \frac{1}{(2 x + 1) ^{3} } } \, dx =$
u = 2 x + 1, d u = 2 d x, d x = d u / 2
$36*1/2 \int\limits^1_0 {u ^{-3} } \, du = 18 * t ^{-2}/(-2) = \\ - 9 * t^{-2}= \\ -9 * ( 2 x + 1 ) ^{-2}$
- 9 / ( 2 + 1 )² - ( - 9 / ( 0 + 1 )² =
- 9 / 9 + 9 = - 1 + 9 = 8

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