 Hi and welcome to the session. Let us discuss the calling question. The question says evaluate integral of 3x plus 1 divided by 2x squared minus 2x plus 3 with respect to x. So now begin with the solution. Let i is equal to integral of 3x plus 1 divided by 2x squared minus 2x plus 3 with respect to x. Now let 3x plus 1 is equal to lambda into derivative of 2x squared minus 2x plus 3 plus mu. Now this implies 3x plus 1 is equal to lambda into 4x minus 2 plus mu. Now in comparing coefficients of right powers of x equals to 4 lambda and 1 equals to minus 2 lambda plus mu. This implies lambda is equal to 3 by 4 and 1 equals to minus 2 lambda plus mu plus 1 is equal to minus 2 into 3 by 4 plus mu. This implies mu is equal to 5 by 2 is equal to integral of 3 by 4 into minus 2 plus 5 by 2 divided by 2x squared minus 3 with respect to 3 by 4 into integral of minus 2 divided by 2x plus 3 with respect to x plus 5 by 2 into integral of plus 3 with respect to x. Differentiating both sides of this equation with respect to x we get dt by dx equals to 4x minus 2 dt is equal to 4x minus 2 into dx 3 by 2 with respect to z equal to 3 by 4 integral of 1 by t with respect to