OrnaVerum
v 6.30.00
28 Jan 2022
updated 28 Jan 2022

Eqn 11c Help

Differentiating loge {u + √(1+u2)} using the 'function of a function' rule we get

d
du
 loge {u + √(1+u2)}
=
1
{u + √(1+u2)}
d
du
{u + √(1+u2)}
 
=
1
{u + √(1+u2)}
{1 + 2u (½) / √(1+u2)}
 
=
1
{u + √(1+u2)}
{√(1+u2) + u} / √(1+u2)
 
=
1 / √(1+u2)
So the anti-derivative (indefinite integral) of 1 / √(1+u2) is loge {u + √(1+u2)}:
∫ du / √(1+u2) = loge {u + √(1+u2)}