Question

How do you integrate `int_1^3sqrt(1+(2x-1/(8x))^2)dx` ?

Answer 1

Given `int_1^3sqrt(1+(2x-1/(8x))^2)dx`
Simplifying the expression under the parentheses

`int_1^3sqrt(1+(4x^2+1/(64x^2)-1/2))dx`
`=>int_1^3sqrt((64x^2+256x^4+1-32x^2)/(64x^2))dx`
`=>int_1^3sqrt((256x^4+32x^2+1)/(64x^2))dx`
`=>1/8int_1^3sqrt((16x^2+1)^2/(x^2))dx`

Considering positive root

`1/8int_1^3(16x^2+1)/(x)dx`
`=>1/8int_1^3(16x+1/x)dx`

Using standard integrals
`=>1/8|(16x^2/2+ln|x|)|_1^3`
`=>|x^2+1/8(ln|x|)|_1^3`
`=>3^2+1/8ln3-(1^2+1/8ln1)`
`=>8+1/8ln3`