 Hi and welcome to the session. Let's work out the following question. The question says evaluate integral x square divided by x square plus 6x plus 12 dx. Let us start with the solution to this question. Let i be the integral given to us at this integral x square divided by x square plus 6x plus 12 dx. This is equal to integral x square plus 6x plus 12 minus 6x minus 12 divided by x square plus 6x plus 12 dx. Now we take this in one bracket and this in one bracket and we separate these two terms. So we will have integral 1 upon x square plus 6x plus 12. Sorry here we will have integral 1 minus 6x plus 12 divided by x square plus 6x plus 12 dx and this is equal to Now integral of 1 dx is simply x minus 3 into integral 2x plus 4 divided by x square plus 6x plus 12 dx. Now what we have done is we have taken 3 common from the numerator. This is equal to x minus 3 into integral 2x plus 4 plus 2 minus 2 divided by x square plus 6x plus 12 dx. This is equal to x minus 3 into integral 2x plus 6 divided by x square plus 6x plus 12 dx minus 2 into integral dx upon x square plus 6x plus 12. Here also we see that we have taken this with this and then minus 2 with the denominator. This is equal to x minus 3 into log of mod x square plus 6x plus because we know that integral of derivative of a function divided by the function itself is log of mod of function minus 3 into 2 into integral dx upon x square plus 6x. Now 12 can be written as 9 plus 3. This is equal to x minus 3 into log of mod x square plus 6x plus 12 minus 6 into integral dx upon x plus 3 the whole square plus square root of 3 square. This is how we can write this. This is equal to x minus 3 into log of mod x square plus 6x plus 12 minus 6 into 1 by root 3 into tan inverse x plus 3 divided by root 3 plus the constant say c. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.