Hi
I did a calculation on an integral but i get the wrong answer, but the correct answer is elusive to me in terms of how it got the final term.
This is the workings i did:
. . . . .\(\displaystyle \displaystyle \int\, \left(\dfrac{x}{x\, +\, 3}\right)^2\, dx\, =\, \int\, \dfrac{x^2}{(x\, +\, 3)^2}\, dx\)
Long division gives us:
. . . . .\(\displaystyle \displaystyle \int\, \left(1\, -\, \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\right)\, dx\)
Partial-fraction decomposition gives us:
. . . . .\(\displaystyle \displaystyle \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\, =\, \dfrac{A}{x\, +\, 3}\, +\, \dfrac{B}{(x\, +\, 3)^2}\)
. . . . .\(\displaystyle \displaystyle 6x\, -\, 9\, =\, A(x\, +\, 3)\, +\, B\)
. . . . .\(\displaystyle \displaystyle 6x\, -\, 9\, =\, Ax\, +\, (3A\, +\, B)\)
. . . . .\(\displaystyle \displaystyle 6\, =\, A\)
. . . . .\(\displaystyle \displaystyle -9\, =\, 3A\, +\, B\, =\, 18\, +\, B\)
. . . . .\(\displaystyle \displaystyle -27\, =\, B\)
So the integral becomes:
. . . . .\(\displaystyle \displaystyle \int\, \left(1\, -\, \dfrac{6}{x\, +\, 3}\, -\, \dfrac{27}{(x\, +\, 3)^2}\right)\, dx\)
. . . . . . . .\(\displaystyle \displaystyle=\, \int\, 1\, dx\, -\, 6\, \int\, \dfrac{1}{x\, +\, 3}\, dx\, -\, 27\, \int\, \dfrac{1}{(x\, +\, 3)^2}\, dx\)
. . . . . . . .\(\displaystyle \displaystyle =\, x\, -\, 6\, \ln|x\, +\, 3|\, -\, \dfrac{27}{x\, +\, 3}\, +\, C\)
Actual answer:
. . . . .\(\displaystyle \displaystyle \large{\mathbf{\color{#ff8c00}{ \mathit{x}\, -\, 6\, \ln\big|\mathit{x}\, +\, 3\big|\, -\, \dfrac{9}{\mathit{x}\, +\, 3}\, +\, C }}}\)
I cannot see how they get -9 over x+3 but wolfram says its -9 also, so i am clearly making an error here but i cannot see it.
Hope some one can see my error.
I did a calculation on an integral but i get the wrong answer, but the correct answer is elusive to me in terms of how it got the final term.
This is the workings i did:
. . . . .\(\displaystyle \displaystyle \int\, \left(\dfrac{x}{x\, +\, 3}\right)^2\, dx\, =\, \int\, \dfrac{x^2}{(x\, +\, 3)^2}\, dx\)
Long division gives us:
. . . . .\(\displaystyle \displaystyle \int\, \left(1\, -\, \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\right)\, dx\)
Partial-fraction decomposition gives us:
. . . . .\(\displaystyle \displaystyle \dfrac{6x\, -\, 9}{(x\, +\, 3)^2}\, =\, \dfrac{A}{x\, +\, 3}\, +\, \dfrac{B}{(x\, +\, 3)^2}\)
. . . . .\(\displaystyle \displaystyle 6x\, -\, 9\, =\, A(x\, +\, 3)\, +\, B\)
. . . . .\(\displaystyle \displaystyle 6x\, -\, 9\, =\, Ax\, +\, (3A\, +\, B)\)
. . . . .\(\displaystyle \displaystyle 6\, =\, A\)
. . . . .\(\displaystyle \displaystyle -9\, =\, 3A\, +\, B\, =\, 18\, +\, B\)
. . . . .\(\displaystyle \displaystyle -27\, =\, B\)
So the integral becomes:
. . . . .\(\displaystyle \displaystyle \int\, \left(1\, -\, \dfrac{6}{x\, +\, 3}\, -\, \dfrac{27}{(x\, +\, 3)^2}\right)\, dx\)
. . . . . . . .\(\displaystyle \displaystyle=\, \int\, 1\, dx\, -\, 6\, \int\, \dfrac{1}{x\, +\, 3}\, dx\, -\, 27\, \int\, \dfrac{1}{(x\, +\, 3)^2}\, dx\)
. . . . . . . .\(\displaystyle \displaystyle =\, x\, -\, 6\, \ln|x\, +\, 3|\, -\, \dfrac{27}{x\, +\, 3}\, +\, C\)
Actual answer:
. . . . .\(\displaystyle \displaystyle \large{\mathbf{\color{#ff8c00}{ \mathit{x}\, -\, 6\, \ln\big|\mathit{x}\, +\, 3\big|\, -\, \dfrac{9}{\mathit{x}\, +\, 3}\, +\, C }}}\)
I cannot see how they get -9 over x+3 but wolfram says its -9 also, so i am clearly making an error here but i cannot see it.
Hope some one can see my error.
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