 Hi, and welcome to our session. Let us discuss the following question. The question says, integrate the following functions given function as pi times x by x plus 1 into x squared plus 9. Let's now begin with the solution. In this question, we have to integrate the function 5x by x plus 1 into x squared plus 9 with respect to x. Now, this is a proper rational function. And we know that whenever we have integrand as proper rational function, then we resolve this rational function into partial fractions. Let pi x pi x plus 1 into x squared plus 9 is equal to a by x plus 1 plus bx plus c by x squared plus 9. Now here, a, b, and c are constants. And we have to determine the value of these constants. This implies 5x is equal to a into x squared plus 9 plus bx plus c into x plus 1. Now put x equals to minus 1. So we have minus 5 equals to a into 10. This implies a is equal to minus 1 by 2. Now we will put x as 0. So we have 0 equals to a into 9 plus c into 1. Now a is equal to minus 1 by 2. So this implies c is equal to 9 by 2. Now we will put x as 1. So we have 5 equals to a into 10 plus b plus c into 2. This implies 5 is equal to 10 a plus 2b plus 2c. You know the value of a and c. So by substituting values of a and c, we get 5 equals to 10 into minus 1 by 2 plus 2b plus 2 into 9 by 2. Now this implies 5 is equal to minus 5 plus 2b plus 9. This implies 1 is equal to 2b and this implies b is equal to 1 by 2. So value of a is minus 1 by 2, value of b is 1 by 2, and value of c is 9 by 2. However, it's a substitute values of a, b, and c in this expression. So now 5x by x plus 1 into x square plus 9 is equal to minus 1 by 2 into x plus 1 plus 1 by 2x plus 9 by 2 by x square plus 9. This is equal to minus 1 by 2 into x plus 1 plus 1 by 2 into x by x square plus 9 plus 9 by 2 into 1 by x square plus 9. Now integral of 5x by x plus 1 into x square plus 9 with respect to dx is equal to minus 1 by 2 into integral of x plus 1 with respect to x plus 1 by 2 into integral of x by x square plus 9 tx plus 9 by 2 into integral of 1 by x square plus 9 tx. Now to get the first integral, we know that integral of 1 by ax plus b dx is log ax plus b divided by derivative of ax plus b, that is a plus c. Now using this formula, integral of 1 by x plus 1 tx is log x plus 1 plus c1. Now look at the second integral. Now we have 1 by 2 into integral of x square plus 9 tx put square plus 9 as t. Now this implies 2x is equal to dt by dx and this implies x dx is equal to dt by 2. So this is equal to 1 by 4 into integral of dt by t and this is equal to 1 by 4 log t. Now t is equal to x square plus 9, so this is equal to 1 by 4 log of x square plus 9. Now the second one is 9 by 2 into integral of 1 by x square plus 9 tx. This is equal to 9 by 2 into integral of 1 by x square plus 3 square dx. We know that integral of 1 by x square plus a square with respect to x is 1 by a tan inverse x by a plus c. So using this formula, this integral is equal to 1 by 3 tan inverse x by 3 plus c3 is equal to 3 by 2 tan inverse x by 3 plus c3. So now we will add all these integrals integral of 5x by x plus 1 into x square plus 9 with respect to x is equal to minus 1 by 2 into integral of 1 by x plus 1 dx that is minus 1 by 2 log x plus 1 plus integral of 1 by x square plus 9 dx into 9 by 2 and this is equal to 3 by 2 tan inverse x by 3 plus integral of x by x square plus 9 dx into 1 by 2 and this is equal to 1 by 4 into log x square plus 9 plus c. Hence our required answer is minus 1 by 2 log x plus 1 plus 1 by 4 log x square plus 9 plus 3 by 2 tan inverse x by 3 plus c. So this completes the session. Bye and take care.