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