Evaluate the integrals in Exercises 1–14.
∫ (2 dx) / (x³...

Question

Evaluate the integrals in Exercises 1–14.

∫ (2 dx) / (x³ √(x² - 1)), where x > 1

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the integral, the integral of 4 multiplied by d x divided by x to the power of 3, multiplied by the square root of x 2 minus 9, where x is greater than 3. OK. So, it appears for this particular problem, we're ultimately asked to evaluate the integral that is provided to us by the prom itself, where X is greater than 3. So, now that we know that we're ultimately trying to evaluate this integral. What we need to do first, so for our first step is we need to recall and use the substitution formula, so we need to write our substitution formula. So let us state that X is equal to 3 multiplied by c of T. Then that will mean. That D. X by DT will be equal to. 3 multiplied by secant oft multiplied by tangent oft, which will mean that dx is going to be equal to 3 multiplied by secant oft multiplied by tangent of t. DT fantastic. Also, let's make a note in blue. Also, we need to note that the square root of x 2 minus 9 will be equal to the square root of parentheses 3 multiplied by ct of t to the 2 minus 9. Which will be equal to the square root of 9 multiplied by c2 oft minus 9, which will be equal to the square root of 9 multiplied by parentheses of c 2 t minus 1. Which will be equal to the square root of 9 multiplied by tangent square of T, which will be all equal to 3 multiplied by tangent of T. We also need to rewrite in the integrand, so let's say that this is our second step. So we need to rewrite in the integrand, so we need to take the integral of 4 divided by X3, multiplied by the square root of X2 minus 9. DX will be equal to the integral of. 4 divided by parentheses, 3 multiplied by sec of T and then we need to note that parentheses 3 multiplied by c of T will be to the power of 3, and this will be multiplied by parentheses of 3 multiplied by tangent of T, fantastic. And this will be all multiplied by parentheses of 3 multiplied by c can oft multiplied by tangent of T. DT which will be. All equal to the integral of 4 divided by 27 multiplied by c2 TDT, which will be equal to 4 divided by 27 multiplied by the integral of cosine squared of TDT. OK, so we now need to recall and use the identity. And let's write this in blue. We need to recall and use the identity that states that cosine squared of t will be equal to 1 plus cosine of 2 t divided by 2, and we will then need to rewrite our integrand and then simplify. So when we use our identity, we will be able to go ahead and write 4 divided by 27 multiplied by the integral of cosine squared of TDT. Will be equal to 4 divided by 27 multiplied by the interval of 1 plus cosine of 2 oft divided by 2. DT which will be equal to. 2 divided by 27 multiplied by the integral of parentheses of 1 plus cosine of 2 t dt will be equal to 2 divided by 27 multiplied by parentheses of the integral of dt plus 1/2 multiplied by the integral of cosine of 2. T multiplied by d of 2 T. Fantastic. And now we need to integrate, and when we integrate, we will be able to write 2 divided by 27 multiplied by parentheses of the integral of DT plus 1/2 multiplied by the integral of. The cosine of 2t multiplied by d of 2T, which will be equal to 2 divided by 27 multiplied by parentheses of T plus 1/2 multiplied by sin of 2 t. Plus C, which will be equal to drum roll, 2 divided by 27 multiplied by T, plus 2 divided by 27 multiplied by sin of t multiplied by cosine of t. Plus C. And let us make a note that sin of 2 T is equal to 2 multiplied by sin oft multiplied by cosine of T. Fantastic. So now, moving right along here, we need to convert T and sin oft multiplied by cosine of T back to X. So, let's make a note that X will be equal to 3 multiplied by c of T, which will mean that c of T is equal to x divided by 3, which will mean that cosine of T will be equal to 3 divided by x. So that will mean that T is equal to inversec can of. X divided by 3. Fantastic. So we can go ahead and write that sign of T. is equal to the square root of 1 minus cosine square of t, which is equal to the square root of 1 minus 3 divided by x 2, which will be equal to the square root of 1 minus 9 divided by x 2. Which will be equal to the square root of x 2 minus 9 divided by x 2, which will be equal to the square root of x 2 minus 9 divided by x. So therefore, we can write that the integral of 4 multiplied by dx divided by x to the power of 3 multiplied by the square root of x 2. -9 will be equal to 2 divided by 27 multiplied by inverse secant of x divided by 3 plus 2 divided by 27 multiplied by parentheses of the square root of x to the 2 minus 9 divided by x. Multiplied by parentheses of 3 divided by x plus c. Which will be all equal to 2 divided by 27 multiplied by inverse sect of x divided by 3. Plus 2 multiplied by the square root of x 2 minus 9 divided by 9 multiplied by x 2 plus c and this is our correct and final answer hooray, we did it. So going back to look at the prom itself to recap what we've learned, we can safely say that our final answer for the evaluated integral once again is without a doubt. 2 divided by 27 multiplied by inverse c can of x divided by 3 plus 2 multiplied by the square root of x 2 minus 9 divided by 9 multiplied by x 2 plus c. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 1–14.∫ (2 dx) / (x³ √(x² - 1)), where x > 1

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Evaluate the integrals in Exercises 1–14.
∫ (2 dx) / (x³ √(x² - 1)), where x > 1