In Exercises 27–40, use a substitution to change the integral ...

Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ x^2 / √(x^2 - 4x + 5) dx

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 by using an appropriate substitution and known integral formulas. The integral of x 2 divided by the square root of x 2 + 6 X plus 10 d X. OK. So it appears for this particular problem we're given an integral, and we're asked to evaluate this integral using an appropriate substitution and a known integral formula or formulas. So with that said, now that we know what we're ultimately trying to solve for, our first step that we need to take is we need to recall how to use complete the square in the denominator. So we need to take our denominator, which is the square root of X2 + 6 X + 10, and we need to go ahead and write that X2 + 6 X + 10. Is equal to parenthesis x 2 + 6 X. Plus 9. Plus parentheses 10 minus 9, which will be equal to parentheses X +32. So once again, in parentheses X + 3, all to the power of 2. Plus one. OK, we also need to set T is equal to X + 3, and we need to state that DT is equal to D X. Therefore, with this information in mind, the integral will become the integral of x to the power of 2. Divided by the square root of X to the power of 2. Plus 6 X. Plus 10, so 6 X. Plus 10, so once again 6 X plus 10 D X will be equal to the integral of parentheses T minus 3 to the power of 2 divided by the square root of T 2 + 1dt which will be equal to the integral of T2. Minus 60 + 9 divided by the square root of T 2 + 1 DT. OK, so now we need to split our integral up, so we need to take the interval of T 2 divided by the square root of T 2 + 1dt will be equal to the integral of 2 + 1 minus 1, all divided by the square root of T 2 + 1. DT + 9 multiplied by the integral of 1 divided by the square root of T 2 + 1 DT. OK. So for the first term, we need to write that T 2 is equal to parentheses T 2 + 1 minus 1. So, with that in mind, we can go ahead and write that the integral of T 2 divided by the square root of T 2 + 1 DT. Will be equal to the integral. Once again it's gonna be equal to the integral of, so once again the interval of T 2 + 1 minus 1 divided by the square root of T 2 + 1. DT which will be equal to the integral of the square root of T 2 + 1. DT minus the integral of 1 divided by the square root of T to the power of 2. Plus one. DT Fantastic. And note that the DT is outside of our expression. OK. So, let's take a moment here to take a breather. Our next step now is we need to recall our standard antiderivatives. So we need to recall that the integral. Of the square root of a 2 plus X 2 D X will be equal to x divided by 2 multiplied by the square root of a 2 plus x 2, plus a 2 divided by 2 multiplied by the natural log of X plus the square root of a 2 plus x 2. Plus C or C denotes some constant value. So, with that in mind, we can go ahead and write that the interval of X divided by the square root of A 2 plus X 2 DT. Once again, DT will be equal to the square root of a 2 plus x 2 plus C once again where C is some constant value. OK, so with that said, we can go ahead and write that the integral of 1 divided by the square root of a power 2 plus x to the power of 2. DX will be all equal to, so this will be equal to the natural log of parentheses X plus the square root of a 2 plus x 2. Plus C. However, we need to note that in this case, A is equal to A2, which is equal to 1. So therefore, when we put everything together, so putting everything together, we will find that the integral. Of T 2 minus 6 T + 9. Divided by all divided by the square root of T 2 + 1. DT Will be all equal to parentheses T divided by 2 multiplied by the square root of T 2 + 1 + 1/2 multiplied by the natural log of parenthesis T plus the square root of T 2 + 1. Minus the natural log of parenthesis T. Plus the square root of T 2 + 1. Minus 6 multiplied by parenthesis of the square root of T 2 + 1 + 9 multiplied by parentheses the natural log of T plus the square root of T 2 + 1. Plus C. Fantastic, plus C. So, at this stage, we need to combine our like terms, so we need to take the interval of T 2 minus 6 T + 9 divided by the square root of T 2 + 1. DT, which will be equal to parentheses divided by 2 minus 6. Once again in parentheses, multiplied by the square root of T 2 + 1. Plus parentheses 1/2. So it's gonna be plus parentheses 1/2 plus 9. -1. Multiplied by the natural log of parenthesis T plus the square root of T to the power 2 + 1. Plus C, which will be equal to T minus 12 divided by 2 multiplied by the square root of T 2 + 1 + 17 divided by 2 multiplied by the natural log of T plus the square root of T 2 + 1. Plus C, and don't forget your parentheses. Fantastic. So with that said, Our next step, so moving right along here, so we were combining like terms, and our last step is we need to finally, we need to substitute back inner value. T is equal to X + 3. And that the square root of T 2 + 1 is equal to the square root of X2 + 6 X + 10. So we're substituting back our values back into our expression. So, with that said, we need to go ahead and write that the integral of X 2 divided by the square root of X2 + 6 X + 10. DX will be equal to. X minus 9 divided by 2. Multiplied by the square root of, so once again multiplied by the square root of X to the power 2 plus 6 X plus 10. Plus 17 divided by 2 multiplied by the natural log of parentheses X plus 3. Plus the square root of. X to the power of 2 + 6 X plus 10. In parentheses, and don't forget that we need to add a plus C to the very end of this expression. And that's it. We've officially solved for this problem. Hooray, we did it. We found our final answer. Hooray, we did it. So with that said, since we found our final answer, let's box our final answer so our professor or teacher knows what our final answer is. And let's go back to the top to summarize what we've learned. So that will mean that once again our final answer. Without a doubt, is as follows. So it's a long answer, so let's go over it slowly. So our final answer once again is X minus 9 divided by 2, multiplied by the square root of X to the power of 2 + 6 X + 10. Plus 17 divided by 2 multiplied by the natural log of parenthesis of X + 3 plus the square root of X to the power 2 + 6 X + 10 + C. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.∫ x^2 / √(x^2 - 4x + 5) dx

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In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.
∫ x^2 / √(x^2 - 4x + 5) dx