Integration of Irrational Algebraic Function - Practice Questions & MCQ

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

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

Illustration 

Evaluate: $\int \frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{\mathrm{x}+2}}$
Let $\mathrm{I}=\int \frac{dx}{(x+1)\sqrt{x+2}}$
Substitute: $x+2=t^2\Rightarrow dx=2tdt$
$\begin{array}{rl}& \Rightarrow \int \frac{dx}{x+1(\sqrt{x+2})}=\int \frac{2tdt}{(t^2-1)\sqrt{t^2}}=2\int \frac{dt}{t^2-1}\\ & =\frac{2}{2}\log \left|\frac{t-1}{t+1}\right|+C=\log \left|\frac{\sqrt{x+2}-1}{\sqrt{x+2+1}}\right|+C\end{array}$

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

To evaluate this type of integral, put $px+q=1/t$
(c) Integrals of the Form
(i) $\int \frac{1}{({\mathrm{ax}}^2+\mathrm{b})\sqrt{{\mathrm{px}}^2+\mathrm{q}}}dx$

To evaluate this type of integral, put $\mathrm{x}=\frac{1}{\mathrm{t}}$

Integrals of the Form $\int x^m{(a+b{x}^n)}^{p}dx$
Working Rule:
Case I : If $P\in N$, expand using binomial and integrate.
Case II : If $P\in I^{-}$(ie, negative integer), write $x=t^k$ where $k$ is the LCM of $m$ and $n$.

Case III : If $\frac{m+1}{n}$ is an integer and $P\leftrightarrow$ fraction, put $(a+b{x}^n)=t^k$ where $k$ is denominator of the fraction $P$.

Case IV: If $(\frac{m+1}{n}+P)$ is an integer and $P\in$ fraction, $\mathrm{put}(a+b{x}^n)=t^k{x}^n$, where $k$ is denominator of the fraction $P$.

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

Illustration
The value of the integral, $I=\int \frac{dx}{1+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}}$ is
Here $\mathrm{a}=1>0$, therefore make the substitution $\sqrt{x^2+2x+2}=t-x$

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

$\begin{array}{rl}& x^2+2x+2=(t-x{)}^2\\ & x^2+2x+2=t^2-2tx+x^2\\ & 2x+2tx=t^2-2\\ & \Rightarrow x=\frac{t^2-2}{2(1+t)}\Rightarrow dx=\frac{t^2+2t+2}{2(1+t{)}^2}dt\\ & 1+\sqrt{{\mathrm{x}}^2+2\mathrm{x}+2}=1+\mathrm{t}-\frac{{\mathrm{t}}^2-2}{2(1+\mathrm{t})}=\frac{{\mathrm{t}}^2+4\mathrm{t}+4}{2(1+\mathrm{t}{)}^2}\end{array}$

Substituting into the integral, we get
$I=\int \frac{2(1+t)(t^2+2t+2)}{(t^2+4t+4)2(1+t{)}^2}dt=\int \frac{(t^2+2t+2)dt}{(1+t)(1+2{)}^2}$

Using the partial fraction,

$\frac{t^2+2t+2}{(t+1)(t+2{)}^2}=\frac{A}{t+1}+\frac{B}{t+2}+\frac{C}{(t+2{)}^2}$
we get $\mathrm{A}=1,\mathrm{B}=0$ and $\mathrm{C}=-2$
$\frac{t^2+2t+2}{(t+1)(t+2{)}^2}=\frac{1}{t+1}-\frac{2}{(t+2{)}^2}$

Hence,
$\begin{array}{rl}\int \frac{{\mathrm{t}}^2+2\mathrm{t}+2}{(1+\mathrm{t})(1+2{)}^2}\mathrm{dt}& =\int \frac{\mathrm{dt}}{\mathrm{t}+1}-2\int \frac{\mathrm{dt}}{(\mathrm{t}+2{)}^2}\\ & =\ln |\mathrm{t}+1|+\frac{2}{\mathrm{t}+2}+\mathrm{C}\end{array}$

Now, put the value of $t$ in terms of $x$
$I=\ln (x+1+\sqrt{x^2+2x+2})+\frac{2}{x+2+\sqrt{x^2+2x+2}}+C$