Verified answer

   4

 = ∫ (1  –  1 ⁄ x²) • (x²  +  1 ⁄ x) dx ... expand

   1

   4

 = ∫ [ x² + (1 ⁄ x) – 1 – (1 ⁄ x³) ] dx ... integrate each term

   1

 = (x³ ⁄ 3)  +  ln[x]  –  x  +  (1 ⁄ (2x²)) ... evaluated [1, 4]

 = (4³ ⁄ 3)  +  ln[4]  –  4  +  (1 ⁄ 32) – [ (1³ ⁄ 3) +  ln[1]  –  1  +  (1 ⁄ (2x²)) ]

 = 18.75 + 0.1667

 = 18.92

No it's not good, you can only integrate term by term when you are adding (or subtracting the terms) not multiplying.

You have to multiply the binomials before integrating.

The integrand is:

x^2 + 1/x - 1 - 1/x^3 dx

x^2 + 1/x -1 - x^-3 dx

=> x^3/3 + ln|x| - x + 1 / (2x^2) | from 1 to 4