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#