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Indefinite Integral: Evaluate each of the following integrals.

âˆ«(12x)/(9x^2) dx

âˆ«(12Vt^3) dt

âˆ«[22(3âˆš(1+4x)) ds] / (2y^3 + 6y^2) fp y^2

âˆ«(1/(6x + dx)) / (c^3x)

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Evaluate each integral.$$\int \frac{1}{x^{2}-6 x+13} d x$$

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Transcript

Hello. I’m your teacher for the question integral 1 to 6 to Alex cubed minus of nine X square plus of two with respect to work. So we need to integrate the following function. How we will integrate this just say 12 X. You minus of minus X square plus of two with respect to X. So first of all you must be knowing the formula that integration of X rays to power and with respect to X can be written as x rays to power and plus one divided by N plus one plus C. Right see is needed when indefinite integral are to be sold. But here we have definite integral. So we will not be using this year. So I’ll be writing simply 12 extras to power four x 4. Then you will be writing ahead nine X cubed by three express to power two plus 1/2 plus one. Right plus two X. And then the upper limit and the global limit. So you will be writing this way. Now moving ahead what we can write that is 4th Resort 12. Access to powerful minus of three threes are nine x Squire plus off two acts as it is 126. So I’ll be writing three times of first of all I’ll be writing a problem it here X cube was there? Right So there should be an X cube. Right so excuse will be a plus two times six minus of three times of one race to power four minus of three times of one Q plus of two into one. Right? So what I’ll be writing ahead is that is three times of Six arrests to power four can be written as 1296 -3 times of 216 Plus of 2. 6 or 12- of 3 -3 plus two. So what I’ll be writing it is that is 36, 18 8139 27 and 1 28 8. And then to 12 to 36 23 triple eight Then minus of 16 threes are 48, 2006 plus 12- of 3 3. And then two. So what I’ll be writing areas that is 388. Right? So eight minus 8 30 0. Then eight minus four is 4 38 minus six is 32. Then 12 minus two will be simply did 10 and that is 3 to 50 will be here. So this is the answer for this. First integral. I’ll be solving the next integral. Also. So just say that the next integral is 1-4. Eight over root t minus of 12. Route over tears to power three with respect to T. So I’ll be writing 1248 T raise to power minus one by two minus of 12. T raise to power three by two with respect to a T. Now see we need to see the calculation here. So I’ll be writing the calculation part also minus of one by two plus one. Then minus of 12 tds two per three by two plus 13 by two plus one. And the upper limit. Level limit as 1-4. Right now I’ll be writing that positive one by two. So it will be at 2016 and then t raise to power positive one by two minus of there will be a five by two. So 24 by five The rise to power five x 2. And then and after 1- four. Right, So just say that I’ll be writing that 16 times of four rest apart on my two. That is Square root of four. That is to so 16 to 32- of 24 x five. And then after four can be written as a two square. So I’ll be writing to disc warrior and then five by two and then minus of I’ll be writing 16 minus of 24 by five because there will be one only everywhere. Right at the place of the So I’ll be writing that two and two being canceled then tourist to power…

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