# Advanced Examples On Integration Set-2

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Consider now an integral of the form \begin{align}\frac{{{P_n}\left( x \right)}}{{\sqrt {Q\left( x \right)} }}.\end{align} We assert that a polynomial $${P_{n - 1}}\left( x \right)$$ can be found such that

$\int {\frac{{{P_n}\left( x \right)}}{{\sqrt {Q\left( x \right)} }}dx = {P_{n - 1}}} \left( x \right)\sqrt {Q\left( x \right)} + \alpha \int {\frac{1}{{\sqrt {Q\left( x \right)} }}} \,\,dx$

where $$\alpha$$ is a constant to be determined. You are urged to convince yourself of the truth of this assertion (by differentiating both sides). Find whether this assertion also holds trues if $$n \le 2.$$ Try solving example 51 (a) using this approach.

Example - 52

Evaluate  \begin{align}\int {\frac{{{x^3} + 2{x^2} + x - 7}}{{\sqrt {{x^2} + 2x - 3} }}} \,dx\end{align}

Solution: We can find a quadratic polynomial $$a{x^2} + bx + c$$ and a constant $$\alpha$$ such that

$\int {\frac{{{x^3} + 2{x^2} + x - 7}}{{\sqrt {{x^2} + 2x - 3} }}} = \left( {a{x^2} + bx + c} \right)\sqrt {{x^2} + 2x - 3} + \alpha \int {\frac{1}{{\sqrt {{x^2} + 2x - 3} }}} \,dx$

Differentiaing both sides, we obtain

\begin{align}&\qquad\frac{{{x^3} + 2{x^2} + x - 7}}{{\sqrt {{x^2} + 2x - 3} }} = \frac{{\left( {a{x^2} + bx + c} \right)\left( {x + 1} \right)}}{{\sqrt {{x^2} + 2x - 3} }} + \left( {2ax + b} \right)\sqrt {{x^2} + 2x - 3} + \frac{\alpha }{{\sqrt {{x^2} + 2x - 3} }}\\&\Rightarrow \quad {x^3} + 2{x^2} + x - 7 = \left( {a{x^2} + bx + c} \right)\left( {x + 1} \right) + \left( {2ax + b} \right)\left( {{x^2} + 2x - 3} \right) + \alpha\end{align}

comparing coefficients on both sides, we obtain

$a = \frac{1}{3},\,\,\,b = \frac{1}{6},\,\,c = \frac{5}{2},\,\,\alpha = - 9$

All that now remains to do is evaluate the integral

$I = \int {\frac{1}{{\sqrt {{x^2} + 2x - 3} }}} \,\,dx = \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} - 4} }}} \,\,dx\,\,\text{which is of the standard form }$

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If we encounter an integral of the form $$R\left( {x,{x^{{a_1}/{b_1}}}, {x^{{a_a}/{b_2}}},.......{x^{{a_n}/{b_n}}}} \right),$$ we can convert this into an entirely rational expression using the substitution $$x = {t^P}$$ where P is the l.c.m of $${b_1},\,\,{b_2}........{b_n}.$$

This substitution will become clear through an example:

Example - 53

Evaluate \begin{align}\int {\frac{{x + \sqrt{{{x^2}}} + \sqrt{x}}}{{x\left( {1 + \sqrt{x}} \right)}}} \,\,dx\end{align}

Solution: We have either cube roots or sixth roots in this expression. The l.c.m of (3, 6) is 6, so that the substitution x = t6 can reduce this expression to an entirely rational form:

\begin{align} & \qquad\quad x={{t}^{6}} \\ & \Rightarrow\quad dx=6{{t}^{5}}dt \\&\;\;\Rightarrow \quad I=6\int{\frac{{{t}^{6}}+{{t}^{4}}+t}{{{t}^{6}}\left( 1+{{t}^{2}} \right)}}.{{t}^{5}}dt \\&\qquad \quad\;\;= 6\int {\frac{{{t^5} + {t^3} + 1}}{{1 + {t^2}}}} \,dt\\ &\qquad\quad\;\;= 6\left\{ {\int {{t^3}dt + \int {\frac{1}{{1 + {t^2}}}} } \,dt} \right\}\\&\qquad\quad\;\;= \frac{3}{2}{t^4} + 6{\tan ^{ - 1}}t + C\\&\qquad \quad\;\;= \frac{3}{2}{t^{2/3}} + 6{\tan ^{ - 1}}\sqrt{x} + C \end{align}

An integral of the form $$R\left( {x,{{\left( {\frac{{ax + b}}{{cx + d}}} \right)}^{\frac{{{p_1}}}{{{q_1}}}}},{{\left( {\frac{{ax + b}}{{cx + d}}} \right)}^{\frac{{{p_2}}}{{{q_2}}}}}........} \right)$$ , can be reduced by the substitution\begin{align}\frac{{ax + b}}{{cx + d}} = {t^m}\end{align}  where m = l.c.m (q1 , q2 ............)

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