 Hi and welcome to the session. Let us discuss the following question. Question says evaluate the following integrals. Given integral is dx upon x square plus 2x plus 5. Here lower limit of the integral is minus 1 and upper limit is 1. Let us now start with the solution. Now we have to find the integral from minus 1 to 1 dx upon x square plus 2x plus 5. First of all let us consider function x square plus 2x plus 5. Clearly we can see this is the denominator of the given integrant. Now this can be written as x square plus 2x plus 1 plus 4. Now these three terms form whole square of x plus 1 and 4 can be written as 2 square. We know a square plus 2ab plus b square is equal to a plus b whole square. So we can write x square plus 2x plus 1 square as whole square of x plus 1 and 4 can be written as square of 2. Now let us consider indefinite integral dx upon x square plus 2x plus 5. Now this integral is equal to integral dx upon whole square of x plus 1 plus square of 2. Now we will solve this integral by substitution method. So we will put x plus 1 is equal to t. Now differentiating both the sides with respect to x we get 1 is equal to dt upon dx. This further implies dx is equal to dt. Now integral dx upon x plus 1 whole square plus 2 square is equal to integral dt upon t square plus 2 square. Now this integral is equal to 1 upon 2 tan inverse t upon 2. Here we have used the formula integral of dx upon x square plus a square is equal to 1 upon a tan inverse x upon a plus c. Now in this integral x has been replaced by t and value of a is equal to 2. Now substituting x plus 1 for t here we get 1 upon t tan inverse x plus 1 upon 2 is equal to integral of dx upon x plus 1 whole square plus 2 square. Now we get definite integral from minus 1 to 1 dx upon x plus 1 whole square plus 2 square is equal to 1 upon 2 tan inverse x plus 1 upon 2 minus 1 to 1. Now this is further equal to 1 upon 2 tan inverse 2 upon 2 minus 1 upon 2 tan inverse 0 upon 2. This is the value of this function at upper limit 1 and this is the value of this function at lower limit minus 1. Now 2 and 2 will get cancelled and we get 1 upon 2 tan inverse of 1 minus 1 upon 2 we know 0 upon 2 is 0 so here we can write tan inverse 0. Now tan inverse 1 is equal to pi upon 4 so we can write pi upon 4 for tan inverse 1 and tan inverse 0 is equal to 0 only so here we can write 1 upon 2 multiplied by 0. Now simplifying further we get pi upon 8 minus 0 which is further equal to pi upon 8. So we get definite integral from minus 1 to 1 dx upon x square plus 2x plus 5 is equal to pi upon 8. This is our required answer this completes the session hope you understood the solution take care and have a nice day.