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Math Help - Integral question

  1. #1
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    Integral question

    These are some questions I can't solve on a practice test, if anyone can help me it'd be great.

    1. \int \frac{5}{x^2+6x+25}dx

    2. \int \frac{3x^2+14x+19}{x^2+4x+5}dx

    For 1. I'm not sure what to do. Just integrating it normally doesn't seem right.
    For 2. I used long division for it and got 3+\frac{2x+4}{x^2+4x+5} and have no idea what to do next.

    And also this one(should be easier)
    \int arctan(x)dx
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  2. #2
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    for the first one u have

    \int\frac{5}{x^2+6x+25}dx=\int\frac{5}{(x+3)^2+16}  dx so let \tan(\theta)=x+3 and im sure u can get it from there.

    for the second one

    we use the same idea used in the first one. \frac{2x+4}{x^2+4x+5}=\frac{2(x+2)}{(x+2)^2+1}.
    once again we let \tan(\theta)=x+2.
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  3. #3
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    Quote Originally Posted by putnam120 View Post
    for the first one u have

    \int\frac{5}{x^2+6x+25}dx=\int\frac{5}{(x+3)^2+16}  dx so let \tan(\theta)=x+3 and im sure u can get it from there.
    Actually there is a simpler way to get that integral (and a more formal ).

    You should have memorized that,
    \int \frac{1}{1+x^2}dx=\tan^{-1} x+C

    Okay, now you complete the square,
    \int \frac{5}{(x+3)^2+16} dx
    Let, u=x+3 thus, u'=1
    Substitution theorem,
    5 \int \frac{1}{u^2+16} du
    Divide through by 16,
    \frac{5}{16} \int \frac{1}{\frac{u^2}{16}+1} du
    Thus,
    \frac{5}{16} \int \frac{1}{\left( \frac{u}{4} \right)^2 +1} du
    Use linear substitution,
    t=u/4 and t'=1/4
    Thus, by substitution theorem,
    \frac{5}{16}\cdot \frac{4}{1} \int \frac{1}{t^2+1}dt
    Thus,
    \frac{5}{4}\tan^{-1} t+C
    Substitute,
    \frac{5}{4}\tan^{-1} (u/4)+C
    Thus,
    \frac{5}{4}\tan^{-1} \left( \frac{x+1}{4} \right)+C
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  4. #4
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    in your last step you mistakenly subed in x+1 for u

    the correct expression is \frac{5}{4}\tan^{-1}\left(\frac{x+3}{4}\right)+C.

    and what you did is basically the same think as making the substitution that i suggested, the only difference is that yours does not require any vast knowledge of trigonometry.
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