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Question

Evaluate the integral 32[2x5(4x5)+2x+5(4x5)]dx

A
772532
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B
774532
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C
762532
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D
772535
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Solution

The correct option is A 772532
Let I=32[2x5(4x5)+2x+5(4x5)]dx
Put 4x5=5t24dx=10tdt
I=527535(52(1+t2)5t+52(1+t2)+5t)tdt
=(52)327535(|t1|+|t+1|)tdt.
=(52)32135(1t)+|(1+t)|tdt+751((t1)+(t+1))tdt
=(52)322135tdt+751t2dt=772532.

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