 Hi, and how are you all today? The question says evaluate integral under root x square plus a square whole upon x dx So let's start with the solution Let us rewrite the integral given to us one again Now here first of all we will be rationalizing the numerator on doing so we have I Equal to integral x square plus a square upon x under root x square plus a square dx Now separating the integral sign we have I is equal to under root x square upon x under root x square plus a square dx Plus integral a square upon x under root x square plus a x dx Now here let us assume This integral as I1 plus this integral as I2 Right now let us find out the answer for the Integral I1 here on multiplying and dividing the numerator by 2 and or simplifying We have in the integral 2x upon under root x square plus a square dx right now here put x square plus a square equal to t Therefore we have 2x dx equal to dt right so now we can write I1 as 1 by 2 In table dt upon under root t which is t raised to the power 1 by 2 On solving this we have 1 by 2 t raised to the power minus 1 by 2 Plus 1 upon minus 1 by 2 plus 1 plus c 1 Which gives us? 1 upon 2 t raised to the power 1 by 2 upon 1 by 2 plus c 1 Which on further simplifying gives us these two will get cancelled Under root t and on substituting back the value of t we have this as The answer to I1 and let this be the first equation Now we need to solve the second integral that we have made that is integral a square dx upon x under root x square plus a square now here let us put x as a tan Theta on doing so we have dx as a second square theta D theta So we have I2 is equal to a square the constant it will be out so we have a second square theta d theta in place of dx upon a tan theta that is the value of x under root a square tan square theta Plus a square which can be simplified now and written as a square second square theta d theta upon tan theta Into a secant Theta because here if we take a square common he will come out and tan square theta plus 1 is secant square theta which then comes out of the square root It become secant theta on further simplifying We have I square as a square Integral in theta upon a tan theta D theta further We have a now this becomes secant theta upon tan theta is what exactly? integral Cosic theta d theta and its integral which is known to us is a log mod Cosic theta minus got theta Plus c2 let this be the second equation and Further on substituting back The value we have a log mod under root a Square plus x square minus a upon x plus c2 and let this be the second equation not this one so on adding the First and the second equation We get I Equal to under root a square plus x square plus a log mod Under root a square plus x square minus a upon x mod close plus C c1 plus t c2 will become C right so this is the final answer for this Question right so hope you understood the whole solution well and enjoyed it to have a nice day ahead