Definite Integral: int[1,3][(x+1)/(x(x^2+1))]dx (I don't get correct value)

Seed5813

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Jan 29, 2018
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Ok so I evaluated the indefinite integral and got the right answer but when I plugged in the bounds, I got the wrong answer.



Calculate the definite integral:

. . . . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, =\, F(x)\, =\, \int_1^3\, \dfrac{x\, +\, 1}{x\, (x^2\, +\, 1)}\, dx\)



Computed by Maxima:

. . .\(\displaystyle \displaystyle \int\, f(x)\, dx\, \)

. . . . . . .\(\displaystyle =\, \ln\left(\big|x\big|\right)\, -\, \dfrac{\ln(x^2\, +\, 1)}{2}\, +\, \arctan(x)\, +\, C\)

. . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, \)

. . . . . . .\(\displaystyle =\, \ln(3)\, -\, \dfrac{\ln(10)}{2}\, +\, \arctan(3)\, +\, \dfrac{2\ln(2)\, -\, \pi}{4}\)

. . . . . . .\(\displaystyle \approx 0.7575409414518656\)




For the last term I had arctan(1) which is exactly double the last term the int calculator had.

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Ok so I evaluated the indefinite integral and got the right answer but when I plugged in the bounds, I got the wrong answer.



Calculate the definite integral:

. . . . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, =\, F(x)\, =\, \int_1^3\, \dfrac{x\, +\, 1}{x\, (x^2\, +\, 1)}\, dx\)



Computed by Maxima:

. . .\(\displaystyle \displaystyle \int\, f(x)\, dx\, \)

. . . . . . .\(\displaystyle =\, \ln\left(\big|x\big|\right)\, -\, \dfrac{\ln(x^2\, +\, 1)}{2}\, +\, \arctan(x)\, +\, C\)

. . .\(\displaystyle \displaystyle \int_1^3\, f(x)\, dx\, \)

. . . . . . .\(\displaystyle =\, \ln(3)\, -\, \dfrac{\ln(10)}{2}\, +\, \arctan(3)\, +\, \dfrac{2\ln(2)\, -\, \pi}{4}\)

. . . . . . .\(\displaystyle \approx 0.7575409414518656\)




For the last term I had arctan(1) which is exactly double the last term the int calculator had.
What did you get when you rearranged your terms to match what the calculator had, and then combined the last two terms into one fraction? ;)
 
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