# Advanced Examples On Integration Set-6

Go back to  'Indefinite Integration'

We now turn our attention to a special class of irrational algebraic functions, which are, though rarely encountered in any exam, are discussed here for the sake of a comprehensive coverage.

These are integrals of the form $$R\left( {x,\,\,\sqrt {Q(x)} } \right)$$ . For example, an integral like \begin{align}\int {\frac{{dx}}{{x + \sqrt {{x^2} - x + 1} }}} \end{align} would fall in this class. Note that uptill now we have not encountered integrals of this form.

A specific substitution scheme described below is used to solve such integrals.

Let $$Q(x) = a{x^2} + bx + c$$

$$\Rightarrow$$ If a > 0, substitute $$\sqrt {a{x^2} + bx + c} = t \pm x\sqrt a$$ .

i.e. $$a{x^2} + bx + c = {t^2} + a{x^2} - 2\sqrt a \,tx$$

$$\Rightarrow$$ If c > 0, substitute $$\sqrt {a{x^2} + bx + c} = tx \pm \sqrt c$$

i.e. $$a{x^2} + bx + c = {t^2}{x^2} + {c^2} - 2\sqrt c \,\,tx$$

$$\Rightarrow$$ If $$a{x^2} + bx + c$$ is factorisable, i.e. it has real roots

$$\alpha \,\,{\rm{and }}\beta$$ such that $$a{x^2} + bx + c = a(x - \alpha )(x - \beta )$$ ,

Substitute $$\sqrt {a{x^2} + bx + c} = (x - \alpha )t$$ .

These substitutions are known as Euler’s substitutions. You will get an idea of why these substitutions work from the following example.

Example – 58

Evaluate the following integrals:

(a) \begin{align}\int {\frac{1}{{x + \sqrt {{x^2} - x + 1} }}} \,\,dx\end{align}

(b) \begin{align}\int {\frac{1}{{x - \sqrt {{x^2} + 2x + 4} }}} \,\,dx\end{align}

Solution: (a) From the Euler substitution scheme, we use the substitution $$\sqrt {{x^2} - x + 1} = t \pm x$$

$$\Rightarrow$$ $${x^2} - x + 1 = {t^2} + {x^2} \pm 2tx$$

$$\Rightarrow$$ $$(1 \pm 2t)x = 1 - {t^2}$$

$$\Rightarrow$$ \begin{align}x = \frac{{1 - {t^2}}}{{1 \pm 2t}}\end{align}

Let us select the “–” sign. Thus,

\begin{align} & dx=\frac{(1-2t)(-2t)-(1-{{t}^{2}})(-2)}{{{(1-2t)}^{2}}}dt \\ &\quad =\frac{3+2{{t}^{2}}}{{{(1-2t)}^{2}}}dt \end{align}

Also,

\begin{align}&x + \sqrt {{x^2} - x + 1} = x + t - x\\\\ &\qquad\qquad\qquad\quad= t\end{align}

Thus, \begin{align}I = \int {\frac{1}{t} \cdot \frac{{3 + 2{t^2}}}{{{{(1 - 2t)}^2}}}\,\,dt} \end{align}

$$\, = \int {\frac{{3 + 2{t^2}}}{{t{{(1 - 2t)}^2}}}\,\,dt}$$

It is obvious that this integral can now be solved using expansion by partial fractions.

You should now understand why the Euler substitutions work. They reduce the integral to a rational from which can be expanded using partial fractions.

(b) The Euler substitution scheme tells us that we can use the following substitution.

\begin{align}&\qquad\quad \sqrt{{{x}^{2}}+2x+4}=t+x\left( \begin{gathered} & \text{We took the positive sign } \\ & \text{since then the denominator } \\ & x-\sqrt{{{x}^{2}}+2x+4}\text{ simply} \\ & \text{becomes -}t \\ \end{gathered} \right) \\ & \Rightarrow \quad {{x}^{2}}+2x+4={{t}^{2}}+{{x}^{2}}+2tx \\ & \Rightarrow \quad x=\frac{4-{{t}^{2}}}{2t-2} \\ & \Rightarrow \quad dx=\frac{1}{2}\frac{(t-1)(-2t)-(4-{{t}^{2}})(1)}{{{(t-1)}^{2}}}\,\,dt \\ & \qquad\qquad=\frac{1}{2}\frac{-{{t}^{2}}+2t-4}{{{(t-1)}^{2}}}\,\,dt \\ & \qquad\qquad=\frac{-({{t}^{2}}-2t+4)}{2\,{{(t-1)}^{2}}}\,\,dt \\ & \Rightarrow \quad I=\int{\frac{1}{-t}\cdot \frac{-({{t}^{2}}-2t+4)}{2{{(t-1)}^{2}}}\,\,dt} \\ & \qquad\;\;\;=\int{\frac{{{t}^{2}}-2t+4}{2t{{(t-1)}^{2}}}\,\,dt} \\ &\qquad\;\;\;=\int{\frac{{{(t-1)}^{2}}+3}{2t{{(t-1)}^{2}}}\,\,dt} \\ & \qquad\;\;\;=\int{\frac{dt}{2t}+\frac{3}{2}\int{\frac{1}{t{{(t-1)}^{2}}}}\,\,dt} \\ \end{align}

The second integral can easily be evaluated using expansion by partial fractions.