Question

Integrals of the form $\int R(x,\sqrt{a{x}^2+bx+c})dx$ are calculated with the aid of one of the three Euler substitutions.
$\text{I.}\sqrt{a{x}^2+bx+c}=t\pm x\sqrt{a}$ if $a>0;$
$\text{II.}\sqrt{a{x}^2+bx+c}=tx\pm \sqrt{c}$ if $c>0;$
$\text{III.}\sqrt{a{x}^2+bx+c}=(x-a)t\pm x$ if $a{x}^2+bx+c=a(x-\alpha )(x-\beta )$ i.e., if $\alpha$ is a real root of $a{x}^2+bx+c=0.$

Which of the following functions does not appear in the primitive of $\frac{dx}{x+\sqrt{x^2-x+1}}$ if $t$ is a function of $x?$

• A. ${\log }_e|t|$
• B. ${\log }_e|t-2|$
• C. ${\log }_e|t-1|$
• D. ${\log }_e|t+1|$

Answer 1

The correct option is B ${\log }_e|t-2|$

$I=\int \frac{dx}{x+\sqrt{x^2-x+1}}$
Since here $c=1>0$, we can apply the second Euler substitution: $\sqrt{x^2-x+1}=tx-1$
$\Rightarrow (2t-1)x=(t^2-1)x^2$
$\Rightarrow x=\frac{2t-1}{t^2-1}$
Substituting into $I$, we get an integral of a rational fraction:
$\int \frac{dx}{x+\sqrt{x^2-x+1}}=\int \frac{-2t^2+2t-2}{t(t-1)(t+1{)}^2}dt$

Now, $\frac{-2t^2+2t-2}{t(t-1)(t+1{)}^2}=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{(t+1{)}^2}+\frac{D}{t+1}$

Integrating both the sides,
$\int \frac{-2t^2+2t-2}{t(t-1)(t+1{)}^2}dt=\int (\frac{A}{t}+\frac{B}{t-1}+\frac{C}{(t+1{)}^2}+\frac{D}{t+1})dt$
$=A{\log }_e|t|+B{\log }_e|t-1|-C\frac{1}{t+2}+D{\log }_e|t+1|+c$