How to solve this? $int_1^3 (x^2-16) /(x-4$

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Answer 1 · 2018-03-17T09:07:51

$int_1^3(x^2-16)/(x-4)dx=12$

$int_1^3(x^2-16)/(x-4)dx$

= $int_1^3((x+4)(x-4))/(x-4)dx$

= $int_1^3(x+4)dx$

= $[x^2/2+4x]_1^3$

= $[3^2/2+4*3-1^2/2-4*1]$

= $9/2+12-1/2-4$

= $12$

How to solve this? int_1^3 (x^2-16) /(x-4int_1^3 (x^2-16) /(x-4