How do I solve this integration by parts question with tan-1?

Question

Using integration of parts, find the exact value of ∫tan-1(1/2x) dx between the limits of 2 and 0.

After I apply the integration by parts rules, I obtain tan-1(x) •x-ln(x2+4) between 2 and 0 however the complete integration they give in the mark scheme is 2θtanθ+2ln(cosθ). Their final answer is 1/2pi-ln2.

Any help in the steps to get here would be very much appreciated!! :)

Answer 1

By D I method

D I

+ tan^-1(x/2) 1

- 2/(x²+4) x

I = xtan^-1(x/2)|²₀ - ∫²₀2x/(x²+4)dx = π/2 - ∫⁸₄du/u = π/2 - (ln(8)-ln(4)) = π/2 - ln(2)

Answer 2

did you mean as written

integral of inverse tangent of 1/2x

or of inverse tangent of (1/2)x?

2nd leads to

xarctan(x/2) - ln(x^2+4)

evaluated between 0 and 2

= 2arctan(2/2) -ln8 - (0 -ln4)

= 2pi/2 - ln8 +ln4

= pi/2 -ln(8/4)

= pi/2 -ln2

or if

integral of arctan(1/(2x)) then

= ln(4x^2 +1)/4 +xarc(1/(2x))

evaluated between 2 and 0

=(ln17)/4 +2arctan(1/4) - (0 +0)

= about .72 + .04pi

= .84