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Math Help - Intergration help

  1. #1
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    Question Intergration help

    1. (5x + 3) / (x^(2) + 2x + 5)

    2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

    i am struggling to intergrate these two. solution method would be appreciated alot.
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  2. #2
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    Quote Originally Posted by bobguy View Post
    1. (5x + 3) / (x^(2) + 2x + 5)

    2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

    i am struggling to intergrate these two. solution method would be appreciated alot.
    \int{\frac{5x + 3}{x^2 + 2x + 5}\,dx} = \frac{5}{2}\int{\frac{2x + \frac{6}{5}}{x^2 + 2x + 5}\,dx}

     = \frac{5}{2}\int{\frac{2x + 2 - \frac{4}{5}}{x^2 + 2x + 5}\,dx}

     = \frac{5}{2}\int{\frac{2x + 2}{x^2 + 2x + 5}\,dx} - 2\int{\frac{1}{x^2 + 2x + 5}\,dx}

     = \frac{5}{2}\int{\frac{2x + 2}{x^2 + 2x + 5}\,dx} - 2\int{\frac{1}{(x + 1)^2 + 4}\,dx}.

    Solve the first integral using a u substitution, and solve the second integral using a trigonometric substitution x + 1 = 2\tan{\theta}.
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  3. #3
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    Quote Originally Posted by bobguy View Post
    1. (5x + 3) / (x^(2) + 2x + 5)

    2. (3x + 6) / [x^(2) + 8x + 7]^(1/2)

    i am struggling to intergrate these two. solution method would be appreciated alot.
    \int{\frac{3x + 6}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} = \frac{3}{2}\int{\frac{2x + 4}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}

     = \frac{3}{2}\int{\frac{2x + 8 - 4}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}

     = \frac{3}{2}\int{\frac{2x + 8}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} - 6\int{\frac{1}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx}

     = \frac{3}{2}\int{\frac{2x + 8}{(x^2 + 8x + 7)^{\frac{1}{2}}}\,dx} - 6\int{\frac{1}{[(x + 4)^2 - 9]^{\frac{1}{2}}}\,dx}

    Solve the first integral using a u substitution and solve the second integral using the substitution x + 4 = 3\cosh{t}.
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