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Question

Evaluate the following integrals
(a) 311(x2+6x+5)dx
(b) 31(x2+6x+5)dx

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Solution

(a)311[(x+3)222]dx

12(2)ln(x+32x+3+2)31

Formula : 1x2a2dx=12aln(xax+a)

14ln⎜ ⎜ ⎜4826⎟ ⎟ ⎟

14ln(32)

(b)(x33+6x22+5x)31

(263+3(8)+5(2))

=(703)

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