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# How do you integrate : $\int \dfrac{x}{{\sqrt {{x^2} + 9} }}dx$ ?

Last updated date: 13th Jul 2024
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Hint: To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value.

The given expression:
$\int \dfrac{x}{{\sqrt {{x^2} + 9} }}dx$
We can integrate this expression by the substitution-
Let, ${x^2} + 9 = t$ .
Now, differentiate the above assumed equation:
$\Rightarrow \dfrac{{d{x^2}}}{{dx}} + \dfrac{d}{{dx}}(9) = \dfrac{{dt}}{{dx}}$
$\Rightarrow 2x + 0 = \dfrac{{dt}}{{dx}}$
$\Rightarrow 2x.dx = dt$
$\Rightarrow x.dx = \dfrac{{dt}}{2}$
Now, use the above equation in the main expression:
$\because \int \dfrac{x}{{\sqrt {{x^2} + 9} }}dx \;$
put $\dfrac{{dt}}{2}$ instead of $x.dx$ .
$= \int \dfrac{{dt}}{2}\dfrac{1}{{\sqrt t }} \\ = \dfrac{1}{2}\int {t^{ - \dfrac{1}{2}}}dt \\ = \dfrac{1}{2}\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C \\ = \sqrt t + C \;$
Now, substitute the actual value of $t$ :
$= \sqrt {{x^2} + 9} + C$
So, the correct answer is “ $= \sqrt {{x^2} + 9} + C$ ”.

Note: Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.