Given,
∫211x(1+x2)dxWe can write 1x(1+x2) as 1x−x1+x2
∴∫211x(1+x2)dx=∫21(1x−x1+x2)dx
Now,
∫21(1x−x1+x2)dx=∫211xdx−∫21x1+x2dx
So, ∫211xdx=[ln|x|]21+C1
for, ∫21x1+x2dx,
Let, u=1+x2⇒dudx=2x⇒du=2xdx⇒12du=xdx
Substituting these values we get,
∫21x1+x2dx=∫2112udu=12∫211udu=12[ln|u|]21+C2
∴∫211xdx−∫21x1+x2dx=[ln|x|]21−12[ln|u|]21+C=[ln|x|]21−12[ln∣∣1+x2∣∣]21+C=[ln2−ln1]−12[ln(1+22)−ln(1+12)]+C=[ln2−ln1]−12[ln5−ln2]+C=ln(21)−12ln(52)+C∴∫211x(1+x2)dx=ln(21)−12ln(52)+C.