Examples On Integration By Substitution Set-10

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Now suppose that we have an integral of the form \(\begin{align}\int {\frac{{L\left( x \right)}}{{Q\left( x \right)}}} \,dx\;or\;\int {\frac{{L\left( x \right)}}{{\sqrt {Q\left( x \right)} }}} \,dx,\end{align}\)  where L(x) is a linear term of the form (px + q) while Q(x) is a quadratic term of the form \(a{x^2} + bx + c.\)

In such a scenario, we express the linear factor L(x) in terms of the derivative of the quadratic factor Q(x) , i.e in terms of Q'(x). This means that we have to find constants \(\alpha\; and\;\beta \) such that

\[L\left( x \right) = \alpha Q'\left( x \right) + \beta \]

\[or\]

\[\begin{align}\left( {px + q} \right) &= \alpha \left( {2ax + b} \right) + \beta \\  &= 2a\alpha x + \left( {\alpha b + \beta } \right)...{\rm{ }}\left( 1 \right)\end{align}\]

Convince yourself from (1) above that we can always find the appropriate \(\alpha\; and\;\beta \) . To actually determine \(\alpha\; and\;\beta \) , we compare both sides of (1). Thus,

\[\begin{align}&\alpha  = \frac{p}{{2a}}\\& \beta = q - \alpha b = q - \frac{{pb}}{{2a}}\end{align}\]

How does all this help?

Consider \(\begin{align} I = \int {\frac{{L\left( x \right)}}{{Q\left( x \right)}}} \,\,dx.\end{align}\) We find \(\alpha \,\,{\rm{and}}\,\,\beta \) such that \(L\left( x \right) = \alpha Q'\left( x \right) + \beta .\) Thus,

\[\begin{align}&&&& I = \int {\frac{{\alpha Q'\left( x \right) + \beta }}{{Q\left( x \right)}}} \,\,dx\end{align}\\ \qquad\qquad\qquad\qquad\qquad= \alpha \int {\frac{{Q'\left( x \right)}}{{Q\left( x \right)}}} \,\,dx + \beta \int {\frac{1}{{Q\left( x \right)}}} \,\,dx \\ = {I_1}{\rm{ }} + {\rm{ }}{I_2}\]

\({I_1}\) can be solved using the substitution \(Q\left( x \right) = t.\) This gives \(Q'\left( x \right)dx = dt\) and\({I_1}\) reduces to\(\begin{align} \alpha \int {\frac{{dt}}{t}} = \alpha \ln \left| t \right|.{I_2}\end{align}\)  can either be expressed in the form of (17) or (21) depending on Q(x) and can subsequently be evaluated

Similarly, consider\[\begin{align}& I = \int {\frac{{L\left( x \right)}}{{\sqrt {Q\left( x \right)} }}} \,\,dx \\ &\;\;= \int {\frac{{\alpha Q'\left( x \right) + \beta }}{{\sqrt {Q\left( x \right)} }}} \,\,dx\\ &\;\;= \alpha \int {\frac{{Q'\left( x \right)}}{{\sqrt {Q\left( x \right)} }}} \,\,dx + \beta \int {\frac{1}{{\sqrt {Q\left( x \right)} }}} \,\,dx \\&\;\;= {I_3}{\rm{ }} + {\rm{ }}{I_4} \end{align}\]

\({I_3}\) can be solved using the substitution \(Q\left( x \right) = t\) which reduces\(\begin{align}{I_3}\,\,to\,\,\alpha \int {\frac{{dt}}{{\sqrt t }}} = \frac{2}{3}\alpha {t^{3/2}}\end{align}\) while \({I_4}\) can be expressed in the form of either (15), (22) or (23) depending on Q(x) and can therefore be evaluated.

Let us consider some examples of this type.

Example – 28

Evaluate the following integrals:

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

(b) \(\begin{align}\int {\frac{{3x - 1}}{{\sqrt {{x^2} + 2x + 2} }}} \,\,dx\end{align}\)

Solution: (a) We find constant \(\alpha\; and\;\beta \) such that

\[x + 2 = \alpha \left( {{x^2} + 2x + 2} \right)' + \beta \]

\[\begin{array}{l} = \alpha \left( {2x + 2} \right) + \beta \\ = 2\alpha x + 2\alpha + \beta \end{array}\]

Thus \(\begin{align}\alpha = \frac{1}{2}\,\,{\rm{and}}\,\,\beta = 1\end{align}\)


\[\begin{align} &\Rightarrow\quad I = \int {\frac{{x + 2}}{{{x^2} + 2x + 2}}} \,\,dx \\ &\quad\qquad= \frac{1}{2}\int {\frac{{2x + 2}}{{{x^2} + 2x + 2}}} \,\,dx + \int {\frac{1}{{{x^2} + 2x + 2}}} \,\,dx\\ &\quad\qquad= \frac{1}{2}\int {\frac{{dt}}{t}} {\rm{ + }}\int {\frac{1}{{{{\left( {x + 1} \right)}^2} + 1}}} \,\,dx\\&\qquad\qquad({\rm{where\;}}t = {x^2} + 2x + 2){\rm{ }}\\&\quad\qquad = \frac{1}{2}\ln \left| t \right| + {\tan ^{ - 1}}\left( {x + 1} \right) + C\\ &\quad\qquad= \frac{1}{2}\ln \left( {{x^2} + 2x + 2} \right) + {\tan ^{ - 1}}\left( {x + 1} \right) + C\\{\rm{ }}\end{align}\]
 

(b) Again, assume \(\alpha\; and\;\beta \) such that

\[3x - 1 = \alpha \left( {{x^2} + 2x + 2} \right)' + \beta \]

\[ = 2\alpha x + 2\alpha + \beta \]

Thus, \(\begin{align}\alpha = \frac{3}{2}\,\,{\rm{and}}\,\,\beta = - 4\end{align}\)

\[\begin{align}& \Rightarrow\quad I = \int {\frac{{3x - 1}}{{\sqrt {{x^2} + 2x + 2} }}} \,\,dx \\ &\quad\qquad= \frac{3}{2}\int {\frac{{2x + 1}}{{\sqrt {{x^2} + 2x + 2} }}} \,\,dx - 4\int {\frac{1}{{\sqrt {{x^2} + 2x + 2} }}} \,\,dx\\ \\ &\quad\qquad= \frac{3}{2}\int {\frac{{dt}}{{\sqrt t }}} \,\ - 4\int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + 1} }}} \,\,dx\\&\qquad\qquad({\rm{where}}\,t = {x^2} + 2x + 2)\\ &\quad\qquad= 3{t^{1/2}} - 4\ln \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 2} } \right| + C\\& \quad\qquad= 3\sqrt {{x^2} + 2x + 2} - 4\ln \left| {\left( {x + 1} \right) + \sqrt {{x^2} + 2x + 2} } \right| + C\end{align}\]

Assume now that we have to integrate an expression of the form \(\begin{align}\frac{{P\left( x \right)}}{{Q\left( x \right)}}\end{align}\)  where Q(x) is a quadratic polynomial while P(x) is a polynomial with degree \(n \ge 2.\) In such a case, we can divide \(P\left( x \right)\,\,{\rm{by}}\,\,Q\left( x \right)\) to obtain a quotient Z(x) and a remainder R(x) whose degree is less than 2:

\[\frac{{P\left( x \right)}}{{Q\left( x \right)}} = Z\left( x \right) + \frac{{R\left( x \right)}}{{Q\left( x \right)}}\]

The right hand side can now be integrated easily.

Think about how we will integrate an expression of the form \(\begin{align}\frac{{P\left( x \right)}}{{\sqrt {Q\left( x \right)} }}\end{align}\) where P(x) and Q(x) have already been described. You will find the answer to this question later in the chapter.

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Download SOLVED Practice Questions of Examples On Integration By Substitution Set-10 for FREE
Indefinite Integration
grade 11 | Questions Set 1
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grade 11 | Answers Set 1
Indefinite Integration
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