Now we need to get set up for the substitution. In other words, we need so solve for \(w\) and get \(dw\).

\[w = 1 - {u^2}\hspace{0.5in} \Rightarrow \hspace{0.5in}dw = - 2u\,du\] Show Step 3

Doing the substitution gives,

\[\int{{\frac{1}{{w + 2\sqrt {1 - w} + 2}}\,dw}} = \int{{\frac{1}{{1 - {u^2} + 2u + 2}}\,\left( { - 2u} \right)du}} = \int{{\frac{{2u}}{{{u^2} - 2u - 3}}\,du}}\] Show Step 4

This integral requires partial fractions to evaluate. Let’s start with the form of the partial fraction decomposition.

\[\frac{{2u}}{{\left( {u + 1} \right)\left( {u - 3} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 3}}\]

Setting the coefficients equal gives,

\[2u = A\left( {u - 3} \right) + B\left( {u + 1} \right)\]

Using the “trick” to get the coefficients gives,

\[\begin{align*}{u = 3:} & \hspace{0.25in} & 6 = & \, 4B\\{u = - 1:} & \hspace{0.25in} & - 2 = & - 4A\end{align*}\hspace{0.25in} \Rightarrow \hspace{0.25in}\begin{aligned} & {A = \frac{1}{2}}\\ & {B = \frac{3}{2}}\end{aligned}\]

The integral is then,

\[\int{{\frac{{2u}}{{\left( {u + 1} \right)\left( {u - 3} \right)}}\,du}} = \int{{\frac{{\frac{1}{2}}}{{u + 1}} + \frac{{\frac{3}{2}}}{{u - 3}}\,du}} = \frac{1}{2}\ln \left| {u + 1} \right| + \frac{3}{2}\ln \left| {u - 3} \right| + c\] Show Step 5

The last step is to now do all the back substitutions to get the final answer.

\[\int{{\frac{1}{{w + 2\sqrt {1 - w} + 2}}\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\ln \left| {\sqrt {1 - w} + 1} \right| + \frac{3}{2}\ln \left| {\sqrt {1 - w} - 3} \right| + c}}\]