JEE Main 2025 Syllabus: Detailed Physics, Chemistry and Mathematics Syllabus

Integration of Irrational Algebraic Function - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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13 Questions around this concept.

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MEDIUM

The value of the integral $\int \frac{d x}{(x-2)^{7 / 8}(x+3)^{9 / 8}}$ is equal to.

$\frac{8}{5}\left(\frac{x-2}{x+3}\right)^{\frac{1}{8}}+C$

$\frac{5}{8}\left(\frac{x-2}{x+3}\right)^{\frac{1}{8}}+C$

$\frac{5}{8}\left(\frac{x+3}{x-2}\right)^{\frac{1}{8}}+C$

$\frac{8}{5}\left(\frac{x+3}{x-2}\right)^{\frac{1}{8}}+C$

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$\int\left(\frac{\ln x-1}{(\ln x)^2+1}\right)^2 d x$ is equal to

$\frac{x}{x^2+1}+C$

$\frac{\ln x-1}{(\ln x)^2+1}$

$\frac{x}{(\ln x)^2+1}+C$

$e^x\left(\frac{x}{x^2+1}\right)+C$

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Concepts Covered - 4

Integration of Irrational Algebraic Function (Part 1)

$\\\mathrm{\mathbf{(a)}\;\;{\mathbf {Integrals \;of \;the \;Form } }}\\\\\mathrm{\;\;\;\;\;\;\;(i)\;\int\frac{1}{(ax+b)\sqrt{px+q}}\;dx}\\\\\mathrm{\;\;\;\;\;\;\;(ii)\;\int\frac{ax+b}{\sqrt{px+q}}\;dx}\\\\\mathrm{\;\;\;\;\;\;(iii)\;\int\frac{\sqrt{px+q}}{{ax+b}}\;dx}\\\\\mathrm{\;\;\;\;\;\;\;(iv)\;\int\frac{1}{(ax^2+bx+c)\sqrt{px+q}}\;dx}$

To evaluate this type of integrals, put px + q = t²

Illustration

$\\\mathrm{Evaluate:\;\;\int \frac{d x}{(x+1) \sqrt{x+2}}}\\\\\text{Let I}=\int \frac{d x}{(x+1) \sqrt{x+2}}\\\\\text { Substitute: } x+2=t^{2} \quad \Rightarrow \quad d x=2 t d t\\\\\Rightarrow \int \frac{d x}{x+1(\sqrt{x+2})}=\int \frac{2 t d t}{\left(t^{2}-1\right) \sqrt{t^{2}}}=2 \int \frac{d t}{t^{2}-1}\\\\=\frac{2}{2} \log \left|\frac{t-1}{t+1}\right|+C=\log \left|\frac{\sqrt{x+2}-1}{\sqrt{x+2+1}}\right|+C$

Integration of Irrational Algebraic Function (Part 2)

$\\\mathrm{\mathbf{(b)}\;\;{\mathbf {Integrals \;of \;the \;Form } }}\\\\\mathrm{\;\;\;\;\;\;\;(i)\;\int\frac{1}{(px+q)\sqrt{ax^2+bx+c}}\;dx}$

To evaluate this type of integral, put px + q = 1/t

$\\\mathrm{\mathbf{(c)}\;\;{\mathbf {Integrals \;of \;the \;Form } }}\\\\\mathrm{\;\;\;\;\;\;\;(i)\;\int\frac{1}{(ax^2+b)\sqrt{px^2+q}}\;dx}$

To evaluate this type of integral,

$\\\mathrm{put\;\;x=\frac{1}{t}}$

Integration by Derived Substitution (Part 2)

$\mathbf { Integrals \;of \;the \;Form } \int x^{m}\left(a+b x^{n}\right)^{p} d x$

Working Rule:

$\\\mathbf { Case \;I: } \text{ If } P \in N, \text { expand using binomial and integrate. }\\\\\mathbf{Case\;II: }\text { If } P \in I^{-} \text {(ie, negative integer), write } x=t^{k}\;\text { where }\\ \mathrm{\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\; k \text { is the LCM of } m \text { and } n.\\\\\mathbf{Case\;III: }\text { If } \frac{m+1}{n} \text { is an integer and } P \leftrightarrow \text { fraction, put }\left(a+b x^{n}\right)=t^{k}\\ \mathrm{\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\; \text { where } k \text { is denominator of the fraction } P.\\\\\mathbf{Case\;IV: }\text { If }\left(\frac{m+1}{n}+P\right) \text { is an integer and } P \in \text { fraction, }\\ \mathrm{\;\;\;\;\;\;\;}\;\;\;\;\;\;\;\;\; \operatorname{put}\left(a+b x^{n}\right)=t^{k} x^{n}, \text { where } k \text { is denominator of the fraction }P.$

Integration Using Euler's Substitution

$\\\mathrm{Integration\;of\;the\;form\;\;\mathbf{\int f(x, \sqrt{a x^{2}+b x+c} )\;d x}\;can\;be\;solved}\\\text{using\ one\;of\;the\;Euler's\;substitutions.}\\\\\text{(i)}\;\;\;\;\sqrt{a x^{2}+b x+c}=t \pm x \sqrt{a}, \text { if } a>0\\\\\text{(ii)}\;\;\;\sqrt{a x^{2}+b x+c}=t x+\sqrt{c}, \text { if } c>0\\\\\text{(iii)}\;\;\;\sqrt{a x^{2}+b x+c}=(x-\alpha) t, \text { If } \alpha \text { is real root of }ax^2+bx+c.$

Illustration

$\\\mathrm{The\;value\;of\;the\;integral,\;\;I=\int \frac{d x}{1+\sqrt{x^{2}+2 x+2}}\;\;is }$

Here a = 1 > 0, therefore make the substitution $\sqrt{x^{2}+2 x+2}=t-x$

Squaring both sides of this equality and reducing the similar terms, we get

$\\{x^{2}+2 x+2}=(t-x)^2\\ x^2+2x+2=t^2-2tx+x^2 \\ 2x+2tx=t^2-2$

$\\\Rightarrow \;\quad x=\frac{t^{2}-2}{2(1+t)} \Rightarrow d x=\frac{t^{2}+2 t+2}{2(1+t)^{2}} d t\\\\\mathrm{\;\;\;\;\;\;\;\;\;1+\sqrt{x^{2}+2 x+2}=1+t-\frac{t^{2}-2}{2(1+t)}=\frac{t^{2}+4 t+4}{2(1+t)^{2}}}\\\text{Substituting into the integral, we get}\\\\\mathrm{I}=\int \frac{2(1+\mathrm{t})\left(\mathrm{t}^{2}+2 \mathrm{t}+2\right)}{\left(\mathrm{t}^{2}+4 \mathrm{t}+4\right) 2(1+\mathrm{t})^{2}} \mathrm{dt}=\int \frac{\left(\mathrm{t}^{2}+2 \mathrm{t}+2\right) \mathrm{dt}}{(1+\mathrm{t})(1+2)^{2}}$

Using the partial fraction,

$\\\frac{t^{2}+2 t+2}{(t+1)(t+2)^{2}}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2)^{2}}\\\text{we get A = 1, B = 0 and C = -2}\\\\\frac{t^{2}+2 t+2}{(t+1)(t+2)^{2}}= \frac{1}{t+1}-\frac{2}{(t+2)^{2}}\\\text{Hence,}\\ {\int \frac{\mathrm{t}^{2}+2 \mathrm{t}+2}{(1+\mathrm{t})(1+2)^{2}} \mathrm{dt}=\int \frac{\mathrm{dt}}{\mathrm{t}+1}-2 \int \frac{\mathrm{dt}}{(\mathrm{t}+2)^{2}}} \\ { \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\ln |\mathrm{t}+1|+\frac{2}{\mathrm{t}+2}+\mathrm{C}}$