The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₀^∞ dx / [(x + 1)(x² + 1)]

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of DX divided by X + 2 multiplied by X2 + 4 between the bonds of 0 and infinity. Now to evaluate this improper integral, first we need to decompose our fraction using partial fraction decomposition, and then we can evaluate the limits. So let's decompose the integral into a sum of simpler fractions, OK. So no, given 1 divided by x + 2 multiplied by x2 + 4. We can rewrite it as a divided by x + 2 + BX + C divided by X2 + 4. And if we multiply both sides by x + 2 multiplied by x2 + 4 to clear the denominators, then 1 will be equal to a multiplied by x2 + 4. Plus b x + c multiplied by x + 2. Now let's use this to help us first figure out A, OK? Let's let X be equal to -2. Well, in that case, that means that 1 will be equal to a multiplied by -22 + 4, OK. + 0. Thus, that means 1 will be equal to 8A, and if 8A equals 1, then A is 1/8. Now that we have A, we can use that to help us figure out B and C. If we expand our equation, OK, then 1 is going to be equal to a multiplied by x2. Plus 4A plus B X 2 + 2 BX + C X + 2 C, OK, that's if we were to expand our previous equation, right? Now if we group our terms for our coefficients, then it's going to be A + B multiplied by X2. Plus 2B plus C multiplied by x plus 4A plus 2c and now we can compare our coefficients, OK. So by comparison, starting with our x2 coefficients, notice that there is no x2 term on the left hand side of our equation. So that means from our right hand side, where the coefficient of x2 is a plus b, it will be equal to 0. And since a equals 1/8, then that means b. Will be equal to -1/8 because 1/8 minus 1/8 would give us 0. Next, taking a look at our coefficients for X. OK, for coefficients for X here, notice that there's no x term on the left hand side, so that means the coefficient of X on the right hand side, which is 2B plus C. OK, or rather not X, my apologies for our constant, OK, for our constant. Our constant term here is one. OK And that means that 1 then will be equal to 4A plus 2C. Remember A is 1/8, so that means 4 multiplied by 1/8 plus 2c equals +14 multiplied by 1/8 is 1/2, so 1/2 plus 2c equals 1, which means that c then will be equal to 1/4. So since we have A, B, and C. Then we can show our decomposition, no, OK, which one that means. 1 divided by X + 2 multiplied by X2 + 4. Is going to be equal to 1 divided by 8 multiplied by x + 2, OK. Plus 1/8 of x minus a quarter bx plus c divided by. X2 + 4. I know we can factor out 1/8 to get 1 divided by X + 2 minus X divided by X2 + 4 + 2 divided by X2 + 4. Now we can go ahead and integrate each of our terms. So let's find the antiderivative of each term and then put it back into our improper integral. So integrating each term. Let me put this in red actually. First, starting with the integral of 1 divided by X + 2 with respect to X. Based on what we know, it's going to be equal to the natural log of the absolute value of x + 2 plus some constant that we can call C1, OK. Next, For the integral of x divided by x2 + 4. With respect to x, OK, we can integrate using substitution. So let's let u be equal to x2 + 4, which means the derivative of u with respect to x will be 2 x, so a half of duu equals xDX. No, if that's the case. Then our integral. OK. With respect to X is going to become the integral of 1 divided by 2 UDU OK, or a half of the integral of UDU, OK. Which in this case then. is going to be equal to 12 of ln of the absolute value of u plus some constant c2 which will be equal to 12 of ln of the absolute value of x2 + 4. OK, plus C2, which if we use what we know about logs, we can rewrite this as the natural log of the absolute value of the square root of x2 + 4 plus some constant c2. Finally, for our third term, the integral of 2 divided by x2 + 4 with respect to x, again, we can use the known formula here that the integral of 1 divided by the sum of 2 squares with respect to x equals 1 divided by the constant multiplied by the inverse tangent of x divided by that constant. In other words, here, if we rewrite this as 2. Can multiplied by the integral of 1 divided by 2 squared. Sorry, X 2. Plus 2 squared with respect to x, then the result will be 2 multiplied by 12 multiplied by the inverse tangent of x divided by 2, which equals the inverse tangent of x divided by 2 plus some constant c3. So now that we have the integrals of all three terms, we can combine them to get our anti derivative. F of x, which will be 1/8 of ln of the absolute value of x + 2 minus ln of the square root of x2 + 4 plus the inverse tangent of x divided by 2. And if we use what we know about natural logs, the properties of logs, then this could be simplified further as an eight of ln of the absolute value of x + 2 divided by the square root of x2 + 4, OK. Plus the inverse tangent of x divided by 2. Now where do we go from here? Well, now we can evaluate the improper integral, so. Thus, if we are finding the limit, eh. As b approaches infinity of f of x between the bones of zero and b, really it's going to be an eighth of the limit as b approaches infinity of our expression ln of the absolute value of x + 2 divided by the square root of x2 + 4. OK, plus the inverse tangent of x divided by 2 between the bones of 0 and b. So now we know that. We'll need to apply our upper and lower bones for this problem. So we're going to have an 18 of our limit as B approaches infinity here. Of Ellen. B + 2 Divided by the square root of B2 + 4, OK, plus the inverse tangent. Of be divided by 2, OK. Minus LN of 0 + 2 or 2 divided by the square root of 02 + 4 or the square root of 4 which is 2, OK, plus the inverse tangent of 0 divided by 2. Now we can simplify our second term here, OK, so no, and our and our first term as well, OK, so that's going to be 1/8 of the limit as B approaches infinity, OK. Of, you know what, I could basically just take this term and carry it down here, but no, all we're doing is for our constant terms here, we'll have the natural log of one, OK. Which here that this would be the natural log of one, let me. Remove this here, so this is going to be. The natural log of one. OK, plus the inverse tangent of 0. Now all of this will simplify to 0, our second term here, OK, but for our first term, then we'll be left with 1/8 of the limit, OK, as B approaches infinity. Of the natural log. Of the absolute value of B + 2 divided by the square root of b2 + 4 OK. Plus, let's put this in our bracket here, plus the limit as B approaches infinity of the inverse tangent of B divided by 2. Now where do we go from here? Well, we're finding our limits, right? And now, if we think about it, OK. We can simplify this entire expression. To eventually solve for it to eventually be 1/8 of the natural log of 1, this is what our first term would simplify to, plus 1/2 of pi. And now if that's the case, then this will be 1/8 of 0 plus 11/2 of pi or 1/8 1/2 of pi, which equals 1/16 of pi. So this would be our improper integral. Thanks a lot for watching everyone. I hope this video helped.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₀^∞ dx / [(x + 1)(x² + 1)]

Watch next

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀^∞ dx / [(x + 1)(x² + 1)]