Endi biz almashtirishga tayyorlanishimiz kerak. Boshqacha qilib aytganda, biz \(w\) ni hal qilishimiz va \(dw\) ni olishimiz kerak.

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

O'zgartirishni amalga oshirish beradi,

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

Bu integral baholash uchun qisman kasrlarni talab qiladi. Keling, qisman kasrning parchalanish shaklidan boshlaylik.

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

Koeffitsientlarni teng belgilash,

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

Koeffitsientlarni olish uchun "hiyla" dan foydalanish,

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

U holda integral,

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

Oxirgi qadam, yakuniy javobni olish uchun barcha orqadagi almashtirishlarni bajarishdir.

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