 Hi, and welcome to the session. Let us discuss the following question. The question says, integrate the following functions, given function S1 by x squared plus 1 into x squared plus 4. Let's now begin with the solution. In this question, we have to integrate the function 1 by x squared plus 1 into x squared plus 4 with respect to x. Now, here the integrand consists of even powers of x. So we will first put x squared as y. So let 1 by x squared plus 1 into x squared plus 4 is equal to 1 by y plus 1 into y plus 4. And we'll resolve 1 by y plus 1 into y plus 4 into partial fractions. This is equal to a by y plus 1 plus b by y plus 4. This implies 1 is equal to a into y plus 4 plus b into y plus 1. Now put y as minus 4. So this implies 1 is equal to b into minus 4 plus 1. This implies b is equal to minus 1 by 3. Now put y as minus 1. So by substituting y as minus 1, we get 1 equals to a into minus 1 plus 4. This implies a is equal to 1 by 3. So 1 by y plus 1 into y plus 4 is equal to 1 by 3 into y plus 1 minus 1 by 3 into y plus 4. Now 1 by y plus 1 into y plus 4 is equal to 1 by x squared plus 1 into x squared plus 4, as we have substituted y in place of x squared. So this is equal to 1 by 3 into x squared plus 1 minus 1 by 3 into x squared plus 4. This implies integral of 1 by x squared plus 1 into x squared plus 4 with respect to x is equal to 1 by 3 into integral of 1 by x squared plus 1 with respect to x minus 1 by 3 into integral of 1 by x squared plus 4 with respect to x. We know that integral of 1 by x squared plus a squared with respect to x is equal to 1 by a tan inverse x by a plus c. So using this formula, integral of 1 by x squared plus 1 with respect to x is equal to tan inverse x plus c1. And using the same formula, integral of 1 by x squared plus 4 with respect to x is equal to 1 by 2, as we can write 4 as square of 2 into tan inverse x by 2 plus c2. This is equal to 1 by 3 tan inverse x minus 1 by 6 tan inverse x by 2 plus c, where c is equal to c1 plus c2. Hence, our required answer is 1 by 3 tan inverse x minus 1 by 6 tan inverse x by 2 plus c. So this completes the session. Bye, and take care.