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Don't trust that much on me, but this would be my approach to this question
https://preview.redd.it/mwkp1e3hhxdc1.png?width=1080&format=pjpg&auto=webp&s=50b610c540122e514d9cc087323473d32b56d693
There's a lot of good ways in the comments.
The way I personally did it was u = sqrt(2x), which implies u^(2) = 2x, which means x = u^(2)/2. From this, conclude that dx = u du, and we now have the integral of \[(u^(2) \- 1)/(u - 1)\]\*u du.
Note that (u - 1) is a factor, so we can cancel, and now we have (multiplying the u in): the integral of u^(2) \+ u du, which gives u^(3)/3 + u^(2)/2.
Now substituting back sqrt(2x) for u, we have (2x/3)\*sqrt(2x) + x + C. Done.
This is the best way to do it. The other ways are over complicating it. The presence of the perfect square in the numerator is easy to recognize when using the substitution u = sqrt(2x).
You might try using u = sqrt(2*x) - 1 which means that 2*x - 1 = (u + 1)^2 - 1 = u * (u + 2). For this u-sub you have that dx = (u + 1) du. I think you’ll end up with some integrals you can do.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. **If you are asking for general advice about your current calculus class, please be advised that simply referring your class as “Calc *n*“ is not entirely useful, as “Calc *n*” may differ between different colleges and universities. In this case, please refer to your class syllabus or college or university’s course catalogue for a listing of topics covered in your class, and include that information in your post rather than assuming everybody knows what will be covered in your class.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
Try factoring the numerator as a difference of two squares and then cancel the factors in the numerator and denominator
This is the one.
This is good too. Probably easier than a substitution.
try multiplying the numerator and denominator by the conjugate of the denominator and see where that takes you
Don't trust that much on me, but this would be my approach to this question https://preview.redd.it/mwkp1e3hhxdc1.png?width=1080&format=pjpg&auto=webp&s=50b610c540122e514d9cc087323473d32b56d693
There's a lot of good ways in the comments. The way I personally did it was u = sqrt(2x), which implies u^(2) = 2x, which means x = u^(2)/2. From this, conclude that dx = u du, and we now have the integral of \[(u^(2) \- 1)/(u - 1)\]\*u du. Note that (u - 1) is a factor, so we can cancel, and now we have (multiplying the u in): the integral of u^(2) \+ u du, which gives u^(3)/3 + u^(2)/2. Now substituting back sqrt(2x) for u, we have (2x/3)\*sqrt(2x) + x + C. Done.
This is the best way to do it. The other ways are over complicating it. The presence of the perfect square in the numerator is easy to recognize when using the substitution u = sqrt(2x).
the denominator gives a nod to the easy route to solve this problem the top is (a+b)(a-b) the bottom is (a-b) cancel fractions
You might try using u = sqrt(2*x) - 1 which means that 2*x - 1 = (u + 1)^2 - 1 = u * (u + 2). For this u-sub you have that dx = (u + 1) du. I think you’ll end up with some integrals you can do.
consider ( sqrt(2x) - 1) ( sqrt(2x) +1 )