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What Euler’s substitutions must be used in integrals of the form: \[\sqrt {a{x^2} + bx + c} \]
 $ \pm \sqrt b x,a > 0 $
 $ \pm \sqrt {{a^3}} x,a > 0 $
 $ \pm \sqrt a x,a > 0 $
None of These

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Last updated date: 20th Jun 2024
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Answer
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Hint: In this question, we need to determine the Euler’s substitution that must be used in evaluating the integrals of the type \[\sqrt {a{x^2} + bx + c} \]. Use all the options in Euler’s substitutions methods to get the desired results and check which of the options is best suitable.
Formula Used: As such there is no formula used. Just to get rid of the square root and quadratic equation, we use Euler’s substitutions methods.

Complete step-by-step answer:
For option A: Our assumption is $ a > 0 $ .
Substitute: \[\sqrt {a{x^2} + bx + c} = x\sqrt b + t\]
Squaring both sides,
We get, \[a{x^2} + bx + c = {x^2}b + {t^2} + 2x\sqrt b \]
\[a{x^2} - {x^2}b + bx - 2x\sqrt b = {t^2} - c\]
On taking out x common from left hand side,
We get, \[x\left( {ax - xb + b - 2\sqrt b } \right) = {t^2} - c\]
Hence we cannot solve further as we have to solve x in terms of constants which is not possible.

For option B: Our assumption is $ a > 0 $ .
Substitute: \[\sqrt {a{x^2} + bx + c} = x\sqrt {{a^3}} + t\]
Squaring both sides,
We get, \[a{x^2} + bx + c = {x^2}{a^3} + {t^2} + 2x\sqrt {{a^3}} \]
\[a{x^2} - {x^2}{a^3} + bx - 2x\sqrt {{a^3}} = {t^2} - c\]
On taking out x common from left hand side,
We get, \[x\left( {ax - x{a^3} + b - 2\sqrt {{a^3}} } \right) = {t^2} - c\]
Hence we cannot solve further as we have to solve x in terms of constants which is not possible.

For option C: Our assumption is $ a > 0 $ .
Substitute: \[\sqrt {a{x^2} + bx + c} = x\sqrt a + t\]
Squaring both sides,
We get, \[a{x^2} + bx + c = {x^2}a + {t^2} + 2x\sqrt a \]
Cancelling $ a{x^2} $ on both sides,
We get, \[bx + c = {t^2} + 2x\sqrt a \]
\[bx - 2x\sqrt a = {t^2} - c\]
On taking out x common from left hand side,
We get, \[x\left( {b - 2\sqrt a } \right) = {t^2} - c\]
So, $ x $ comes out to be: \[x = \dfrac{{{t^2} - c}}{{b - 2\sqrt a }}\]
Hence, we calculate x in terms of constants for $ a > 0 $ .
So, option C is correct, that is $ \pm \sqrt a x,a > 0 $ .
So, the correct answer is “Option C”.

Note: Here we use Euler’s substitutions to check whether x comes out to be in constants for $ a > 0 $ .As to get rid of the square roots and quadratic equation. So, we verified all the options to get the desired results. We have three Euler’s substitutions for different values of $ \left( {a,b,c} \right) $ . First we assume $ a > 0 $ , for second we assume $ b \ne 0 $ and third we assume $ c > 0 $ .