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    I have been set the following question
    \int \frac{1}{x^2+2x+5}\ dx.
    How do I go about factorising the x^2+2x+5 as it doesn't seem to factorise.

    Barney
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    (Original post by Barney63)
    I have been set the following question
    \int \frac{1}{x^2+2x+5}\ dx.
    How do I go about factorising the x^2+2x+5 as it doesn't seem to factorise.

    Barney
    Start off by completing the square.
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    HINT:  x^2 + 2x + 5 = x^2 + 2x + 1 + 4
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    Depending on the exam board, after completing the square, you may be able to use some standard formulae given in the formula book which makes this question fairly easy. If not then the question is much longer.
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    Right I have done this question, but get a different answer to Wolfram Alpha.
    I am the same as WA up until  \frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du
    What I have done is let s=\frac{u}{4}
    that giving me \frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C subbing s and u back in \frac {1}{4} tan^-^1 \left( \frac {x+1}{4} \right) +C
    WA on the other hand does let s= \frac {u}{2} and ds = \frac {1}{2} giving \frac {1}{2} tan^-^1 \left( s \right) +C subbing s and u back in \frac {1}{2} tan^-^1 \left( \frac {x+1}{2} \right) +C
    Why?

    Barney
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    (Original post by Barney63)
    Right I have done this question, but get a different answer to Wolfram Alpha.
    I am the same as WA up until  \frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du
    What I have done is let s=\frac{u}{4}
    that giving me \frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C subbing s and u back in \frac {1}{4} tan^-^1 \left( \frac {x+1}{4} \right) +C
    WA on the other hand does let s= \frac {u}{2} and ds = \frac {1}{2} giving \frac {1}{2} tan^-^1 \left( s \right) +C subbing s and u back in \frac {1}{2} tan^-^1 \left( \frac {x+1}{2} \right) +C
    Why?

    Barney
    why have you let s=u/4? that doesn't help you because then s^2 = u^2/16!
    And you can't integrate a function of s with respect to u - you have to convert the whole integral consistently.
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    Sorry I didn't follow that.

    Barney
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    (Original post by Barney63)
    Sorry I didn't follow that.

    Barney
    You are making this much too hard. Why have you taken the factor of \frac{1}{4} out? This has created an unnecessary fractional denominator.
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    \int \frac {1}{(x+1)^2+4}dx then let u=x+1
    \int \frac {1}{u^2+4}du then
    \int \frac{1}{4 \left( \frac {u^2}{4} +1 \right)} du then
    \frac {1}{4} \int \frac {1}{\frac{u^2}{4}+1} du
    Or how would you recommend doing it?

    Barney
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    No substitution is necessary but, if you feel the need to make one, stop here and look at the standard integrals in your formula book.

    (Original post by Barney63)

    \displaystyle \int \frac {1}{(x+1)^2+4}\, dx

    then let u=x+1


    \displaystyle \int \frac {1}{u^2+4}\, du
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    We haven't been given/use a formula book

    Barney
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    (Original post by Barney63)
    Right I have done this question, but get a different answer to Wolfram Alpha.
    I am the same as WA up until  \frac {1}{4} \int \frac {1} {\frac {u^2}{4} +1} du
    What I have done is let s=\frac{u}{4}
    You should to substitute s=\frac{u}{2}
    that giving me \frac{1}{4} \int \frac {1}{s^2 +1} du = \frac {1}{4} tan^-^1 \left( s \right) +C
    1. Your substitution here is wrong totally
    AS you wrote let s be s=\frac{u}{4}. OK.
    then
    u=4s ->u^2=16s^2 ->\frac{u^2}{4}=4s^2->(2s)^2
    and you have to substitute the du factor too.
    Without this, the integral in that form you wrote gives different funtion. Namely
    \frac{1}{4} \int \frac {1}{s^2 +1} du =\frac {1}{s^2 +1}\cdot u+C
    because s is a constant with respect to u. (integration factor is du here)
    THe right method would be:
    u=4s ->ratio of differentials \frac{du}{ds}=4->du=4ds
    \frac{1}{4} \int \frac {1}{(2s)^2 +1} 4ds =\int \frac {1}{(2s)^2 +1}ds
    and now use the t=2s substitution
    from this
    s=\frac{t}{2}->ds=\frac{1}{2}dt
    and
    \frac{1}{2} \int \frac{1}{1+t^2} dt
    which is a base integral for t.
    2. With s=\frac{u}{2} you will get similar form more simple way
    then  u=2s->du=2ds
    \frac{1}{4}\int \frac{1}{1+s^2} \cdot 2\cdot ds=\frac{1}{2}\int \frac{1}{1+s^2} ds
    3. Most simple method without substitution is to use the known rules for
    differentiation, composite function and the chain rule.
    \displaystyle \frac{1}{4} \int \frac{1}{\left (\frac{x+1}{2}\right )^2+1}dx=
    \displaystyle =\frac{1}{4} \cdot \frac{arc tan \left (\frac{x+1}{2}\right )}{\frac{1}{2}}= \frac{1}{2} \cdot arc tan \left (\frac{x+1}{2}\right )
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    (Original post by Barney63)
    We haven't been given/use a formula book

    Barney
    What course are you studying?
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    Its the Institute of Maths Polymaths course, an access course for the maths degree.

    Barney
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    (Original post by Barney63)
    Its the Institute of Maths Polymaths course, an access course for the maths degree.

    Barney
    Ok now I understand why you appeared to be making a meal of this.

    A Level students are provided with a booklet containing this standard integral (along with many others):

    \displaystyle \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a}\tan^{-1}\frac{x}{a}+k
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    So that would give
    \int \frac {1}{u^2+2^2}
    \frac{1}{2}tan^-^1 \frac {u}{2} +C

    Barney
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    (Original post by Barney63)
    So that would give
    \int \frac {1}{u^2+2^2}
    \frac{1}{2}tan^-^1 \frac {u}{2} +C

    Barney
    Yes (if you remember the du at the end of your integral).

    You might like to think about how you could prove this standard result (it isn't as difficult as you might think as you know the answer you are seeking).
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    (Original post by Mr M)
    Ok now I understand why you appeared to be making a meal of this.

    A Level students are provided with a booklet containing this standard integral (along with many others):

    \displaystyle \int \frac{1}{x^2+a^2} \, dx = \frac{1}{a}\tan^{-1}\frac{x}{a}+k
    I have never seen anything like that in my life...
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    Thanks for the help, you have made things much clearer.
    Looking back in my notes we were given that formula in class.
    Once again thank you.

    Barney
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    (Original post by Barney63)
    Thanks for the help, you have made things much clearer.
    Looking back in my notes we were given that formula in class.
    Once again thank you.

    Barney
    Ok ... you are welcome.
 
 
 
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