Ok so integrating by parts I get
[-x / 3(2 + 3x)] + [(1/9) ln (2 + 3x)] + C
and in the book using tables to integrate the answer is
[(1/9) ln (2 + 3x)] + [2/ 9(2 + 3x)] + C
I cant see what im doing wrong.. thnx
[-x / 3(2 + 3x)] + [(1/9) ln (2 + 3x)] + C
and in the book using tables to integrate the answer is
[(1/9) ln (2 + 3x)] + [2/ 9(2 + 3x)] + C
I cant see what im doing wrong.. thnx
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Your answer is totally correct, I checked with Wolfram Alpha and it verifies that your answer differentiates back to the original function given.
Your answer and the answer in the book differ by a constant, which is fine because that is absorbed into the constant of integration.
Integration by parts is really not the best method and you do not need the square brackets as division comes before addition by default.
Your answer and the answer in the book differ by a constant, which is fine because that is absorbed into the constant of integration.
Integration by parts is really not the best method and you do not need the square brackets as division comes before addition by default.
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∫x/(2 + 3x)² dx
1/9*∫x/(2/3 + x)² dx
1/9*∫(x + 2/3 - 2/3)/(x + 2/3)² dx
1/9*∫1/(x + 2/3) dx - 2/27*∫1/(x + 2/3)² dx
1/9*ln|3x + 2| + 2/(27x + 18) + C
1/9*∫x/(2/3 + x)² dx
1/9*∫(x + 2/3 - 2/3)/(x + 2/3)² dx
1/9*∫1/(x + 2/3) dx - 2/27*∫1/(x + 2/3)² dx
1/9*ln|3x + 2| + 2/(27x + 18) + C