Question

Show that \[\int{du \over \sqrt{1 + u^2}} = \ln|u + \sqrt {1 + u^2}| + c\]

Answer 1

I used a table but then was later told that I needed to do it by hand

Answer 2

ok its gonna be done by substitution method

Answer 3

or wait can we do that \[d/du= \ln [u=\sqrt{1+u^2}]=\int\limits_{?}^{?}(1)/\sqrt{1+u^2)}\]

Answer 4

?

Answer 5

Use \[
u=\tan(x)\\
du= \sec^2(x) dx\\
\sqrt{ 1 + u^2} =\sqrt{ 1 +\tan^2(x)}= \sec(x)\\
\frac{du}{\sqrt{ 1 + u^2}}=\frac{\sec^2(x) dx}
{\sec(x)}= \sec(x) dx
\]
You can now finish it from here.