In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Some Advanced Integration Techniques

Go back to  'Indefinite Integration'

In this section, we intend to discuss more advanced techniques of integration and specific substitutions that we’ve not covered till now. This section can be treated as optional and the types of integrals/substitutions discussed in this section will rarely be encountered in an actual exam; for a comprehensive study, however, you can give this section a good reading (but again, don’t try to remember all the formulae)

In the following discussion, the symbol L(x) represents a linear expression in x (of the form ax +b) while Q(x) represents a quadratic expression (of the form ax2 + bx + c). Pn(x) would represent a polynomial of degrees n greater than two. M(x) represents a general polynomial. R(a, b, c.......) would represent a rational function of the variables a, b, c....... .

Recall that we have already developed the requisite techniques to evaluate these types of integrals:

\( \begin{align}(a)\qquad &\frac{{{M_1}\left( x \right)}}{{{M_2}\left( x \right)}}\\&\left( \begin{gathered}{M_2}\left( x \right){\text{is factorisable}}\\{\text{into linear and or}}\\{\text{ quadratic factors}}\end{gathered} \right)\end{align}\) : If deg  M 1 (x) < deg (M2 (x), we expand this expression using partial fractions. If deg (M(x)) deg (M2 (x)), we first divide M1 (x) by M2 (x) to obtain the quotient and the remainder and then apply expansion by partial fractions.
\(\begin{align} (b)\qquad \frac{1}{{Q\left( x \right)}}\end{align}\) : Depending on the coefficients in Q(x), this integral is of the standard form (17) or (21)
 \(\begin{align}(c)\qquad \frac{{L\left( x \right)}}{{Q\left( x \right)}}\end{align}\) :

Find constants \(\alpha \,\,{\rm{and}}\,\,\beta \) such that \(L\left( x \right) = \alpha Q'\left( x \right) + \beta \)

(Q(x) is not factorisable)

\(\begin{align}(d)\qquad \frac{1}{{\sqrt {Q\left( x \right)} }}\end{align}\) : Depending on the coefficients in Q (x), this integral is of the standard form (15), (22) or (23)
  \(\begin{align} (e)\qquad \frac{{L\left( x \right)}}{{\sqrt {Q\left( x \right)} }}\end{align}\) : Find constants \(\alpha \,\,{\rm{and}}\,\,\beta \) such that \(L\left( x \right) = \alpha \,Q'\left( x \right) + \beta \)
\(\begin{align}(f)\qquad\sqrt {Q\left( x \right)} \end{align}\) : Depending on the coefficients in Q (x), this integral is of the standard form (26), (27) or (28)
 \(\begin{align} (g)\qquad L\left( x \right)\sqrt {Q\left( x \right)} \end{align}\) : Find constants \(\alpha \,\,{\rm{and}}\,\beta \) such that \(L\left( x \right) = \alpha Q'\left( x \right) + \beta \)

 

Let us now consider more forms of this sort. You will observe that the basic unifying theme to solve any integral is the same: we must somehow try to reduce the integral given to us to one of the standard simpler forms.

Consider an expression of the form \(\begin{align}\frac{{{Q_1}\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}.\end{align}\) To integrate this, we find constants \(\alpha ,\beta \,\,{\rm{and}}\,\,\gamma \) such that

\[{Q_1}\left( x \right) = \alpha {Q_2}\left( x \right) + \beta {Q_2}'\left( x \right) + \gamma \]

Thus, this integral becomes

\[\begin{align}&I = \alpha \int {\frac{{{Q_2}\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}} \,\,dx\, + \beta \int {\frac{{{Q_2}'\left( x \right)}}{{\sqrt {{Q_2}\left( x \right)} }}} \,dx + \,\gamma \int {\frac{1}{{\sqrt {{Q_2}\left( x \right)} }}} \,dx \\\,\\& = \quad \alpha {I_1} \qquad \qquad +\quad \,\beta {I_2}\qquad + \qquad\qquad \gamma {I_3} \qquad \,\,\,\,\,\end{align}\]

I1 is actually \(\int {\sqrt {{Q_2}\left( x \right)} } \,dx\) which, depending on what the coefficients of Q2(x) are, is one of the standard forms (26), (27) or (28) I2 can be evaluated using the substitution Q2(x) = t.  I3 is again one of the standard forms (15), (22) or (23), depending on the coefficients of Q2(x).

Using the same approach, we can evaluate integrals of the form \(\int {{Q_1}\left( x \right)} \sqrt {{Q_2}\left( x \right)} \,\,dx.\) If we again express Q1(x) in terms of Q2(x) as described above, this integral becomes

\[\begin{align}&I = \alpha \int {{Q_2}\left( x \right)\sqrt {{Q_2}\left( x \right)} } \,dx + \beta \int {{Q_2}'\left( x \right)\sqrt {{Q_2}\left( x \right)} } \,dx + \gamma \int {\sqrt {{Q_2}\left( x \right)} } \,\,dx\\&\,\,\,\, = \,\alpha {I_1}\qquad \,\,\,\, +\qquad \,\beta \,{I_2}\,\,\,\,\,\,\,\, + \qquad\,\,\,\,\,\,\,\,\gamma {I_3}\end{align}\]

How to evaluate the integrals I2 and I3 should be obvious. How to evaluate I1 is discussed in the

following example.

Download SOLVED Practice Questions of Some Advanced Integration Techniques for FREE
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
grade 11 | Answers Set 1
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration
grade 11 | Answers Set 2
Download SOLVED Practice Questions of Some Advanced Integration Techniques for FREE
Indefinite Integration
grade 11 | Questions Set 1
Indefinite Integration
grade 11 | Answers Set 1
Indefinite Integration
grade 11 | Questions Set 2
Indefinite Integration
grade 11 | Answers Set 2
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school