JEE Main & Advanced Mathematics Indefinite Integrals Evaluation of the Various Forms of Integrals by Use of Standard Results

(1) \[\int_{{}}^{{}}{\frac{dx}{a{{x}^{\mathbf{2}}}+bx+c}}\] :

Working rule : We write \[\int_{{}}^{{}}{\frac{dx}{a{{x}^{2}}+bx+c}}\]

\[=\frac{1}{a}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+\frac{b}{a}x+\frac{c}{a}}}\] \[=\frac{1}{a}\int_{{}}^{{}}{\frac{dx}{{{\left( x+\frac{b}{2a} \right)}^{2}}+\frac{c}{a}-\frac{{{b}^{2}}}{4c}}}\], which is of the form \[\int_{{}}^{{}}{\frac{dx}{{{X}^{2}}-{{A}^{2}}},\,\int_{{}}^{{}}{\frac{dx}{{{X}^{2}}+{{A}^{2}}}}}\text{ or }\int_{{}}^{{}}{\frac{dx}{{{A}^{2}}-{{X}^{2}}}}\].

(2) \[\int_{{}}^{{}}{\frac{dx}{\sqrt{a{{x}^{\mathbf{2}}}+bx+c}}}\] : This can be reduced to one of the forms of \[\int_{{}}^{{}}{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}\], \[\int_{{}}^{{}}{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}\] or \[\int_{{}}^{{}}{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}\].

(3) \[\int_{{}}^{{}}{\sqrt{a{{x}^{\mathbf{2}}}+bx+c}\mathbf{ }}dx\] : This can be reduced to one of the forms of \[\int_{{}}^{{}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\,dx,\,\int_{{}}^{{}}{\sqrt{{{x}^{2}}-{{a}^{2}}}\,dx\text{ or }\int_{{}}^{{}}{\sqrt{{{a}^{2}}+{{x}^{2}}}}\,dx}\].

(4) \[\int_{{}}^{{}}{\frac{(px+q)\,dx}{a{{x}^{2}}+bx+c}},\] \[\int_{{}}^{{}}{\frac{(px+q)\,dx}{\sqrt{a{{x}^{2}}+bx+c}}}\], \[\int_{{}}^{{}}{(px+q)\,\sqrt{a{{x}^{2}}+bx+c}}\,dx\] :

For the evaluation of any of these integrals,

Put \[px+q=A\] {differentiation of \[(a{{x}^{2}}+bx+c)\}+B\]

Find A and B by comparing the coefficients of like powers of x on the two sides.

In this way the integral breaks up into two parts.

(5) Integrals of the form  \[\int_{{}}^{{}}{\frac{{{x}^{\mathbf{2}}}+\mathbf{1}}{{{x}^{\mathbf{4}}}+\mathbf{k}{{x}^{\mathbf{2}}}+\mathbf{1}}}\]dx,

\[\int_{{}}^{{}}{\frac{{{x}^{\mathbf{2}}}-\mathbf{1}}{{{x}^{\mathbf{4}}}+k{{x}^{\mathbf{2}}}+\mathbf{1}}}dx\],\[\int_{{}}^{{}}{\frac{dx}{{{x}^{\mathbf{4}}}+k{{x}^{\mathbf{2}}}+\mathbf{1}}}\], where \[k\in R\]

Working Method

(i) To evaluate these types of integrals divide the numerator and denominator by \[{{x}^{2}}\].

(ii) Put \[x+\frac{1}{x}=t\] or \[x-\frac{1}{x}=t\] as required.

(6) \[\int_{{}}^{{}}{\frac{{{x}^{2}}+{{a}^{2}}}{{{x}^{4}}+k{{x}^{2}}+{{a}^{4}}}\,dx,\,\int_{{}}^{{}}{\frac{{{x}^{2}}-{{a}^{2}}}{{{x}^{4}}+k{{x}^{2}}+{{a}^{4}}}}\,dx}\], where \[k\] is a constant, negative or zero.

These integrals can be obtained by dividing numerator and denominator by \[{{x}^{2}}\], then putting \[x-\frac{{{a}^{2}}}{x}=t\] and \[x+\frac{{{a}^{2}}}{x}=t\] respectively.

(7) Substitution for some irrational functions

(i) \[\int_{{}}^{{}}{\frac{dx}{\sqrt{(x-\alpha )\,(x-\beta )}},\,\int_{{}}^{{}}{\sqrt{\left( \frac{x-\alpha }{\beta -x} \right)}}\,dx}\]

\[\int_{{}}^{{}}{\sqrt{(x-\alpha )\,(\beta -x)}\,dx}\], Put \[x=\alpha {{\cos }^{2}}\theta +\beta {{\sin }^{2}}\theta \].

(ii) \[\int_{{}}^{{}}{\frac{dx}{(px+q)\,\sqrt{(ax+b)}}}\], Put \[ax+b={{t}^{2}}\]

(iii) \[\int_{{}}^{{}}{\frac{dx}{(px+q)\,\sqrt{a{{x}^{2}}+bx+c}}}\], Put \[px+q=\frac{1}{t}\]

(iv) \[\int_{{}}^{{}}{\frac{dx}{(p{{x}^{2}}+r)\,\sqrt{(a{{x}^{2}}+c)}}}\], at first \[x=\frac{1}{t}\] and then \[a+c{{t}^{2}}={{z}^{2}}\].

(8) Integrals of the form \[\int{\frac{dx}{P\sqrt{Q}}}\], (where \[\mathbf{P}\] and \[\mathbf{Q}\] are linear or quadratic expressions in \[\mathbf{x}\]) : To evaluate such types of integrals, we have following substitutions according to the nature of expressions of \[P\] and \[Q\] in \[x\] :

(i) When \[Q\] is linear and \[P\] is linear or quadratic, we put \[Q={{t}^{2}}\].

(ii) When \[P\] is linear and \[Q\] is quadratic, we put \[P=\frac{1}{t}.\]

(iii) When both \[P\] and \[Q\] are quadratic, we put \[x=\frac{1}{t}.\]