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 $a{x}^2+bx+c=0.$

$\int \frac{xdx}{\left(\sqrt{7x-10-x^2}\right)}$ can be evaluated by substituting for $x$ as

• A. $x=\frac{5+2t^2}{t^2+1}$
• B. $x=\frac{5-t^2}{t^2+2}$
• C. $x=\frac{2t^2-5}{3t^2-1}$
• D. none of these

Answer 1

The correct option is A $x=\frac{5+2t^2}{t^2+1}$

In this case, $a<0$ and $c<0$.
Therefore, neither the first nor the second Euler substitution is applicable. But the quadratic $7x-10-x^2$ has real roots $\alpha =2,\beta =5.$
Therefore, we use the third Euler substitution:
$\sqrt{7x-10-x^2}=\sqrt{(x-2)(5-x)}=(x-2)t$
$\Rightarrow 5-x=(x-2)t^2$
$\Rightarrow x=\frac{5+2t^2}{t^2+1}$