How to solve this? #int_1^3 (x^2-16) /(x-4# Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Mar 17, 2018 #int_1^3(x^2-16)/(x-4)dx=12# Explanation: #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# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1549 views around the world You can reuse this answer Creative Commons License