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Math Help - Help integrating this simple function.

  1. #1
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    Help integrating this simple function.

    I have this function: f(x) = (4x-3)/(x + 1) (upper limit is 3/4, lower limit is 0)

    i have trouble integrating it; if it were simply f(x) = 4x/(x +1) then that would be easy but the -3 causes problems...

    in my solutions it has a really strange explanation of "argtgx" which i have never seen before, help very very appreciated please!
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  2. #2
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    Rewrite it as \displaystyle \int{\frac{4x}{x^2 + 1} - \frac{3}{x^2 + 1}\,dx}.

    The first term is integrated using a \displaystyle u substitution, while the second term is integrated using the substitution \displaystyle x = \tan{\theta}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Rewrite it as \displaystyle \int{\frac{4x}{x^2 + 1} - \frac{3}{x^2 + 1}\,dx}.

    The first term is integrated using a \displaystyle u substitution, while the second term is integrated using the substitution \displaystyle x = \tan{\theta}.
    I get the first term, but honestly, having done calculus for years I've NEVER seen an integration/differentiation using tan, how is that done? i only use cos and sin...
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  4. #4
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    Make the substitution \displaystyle x = \tan{\theta}. What is \displaystyle \frac{dx}{d\theta}?
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  5. #5
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    note ...

    \dfrac{d}{dx} \arctan{x} = \dfrac{1}{x^2+1}
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