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Math Help - More integration:evaluation

  1. #1
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    More integration:evaluation

    Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

    1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

    I got: x -5/(2x^3) -3/(x^3) -2 +c

    2) integrate: sinx/(squareroot of cosx) dx

    I got: -0.5(squareroot of cosx) +c

    3) integrate (e^x)squareroot(e^x+4) dx
    I got: 0.5(squareroot e^x+4) +c

    4) integrate xe^(x^2+1) dx

    I got: 0.5e^(x^2+1) +c

    5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
    Sorry about this 1 a bit hard to read.
    It is 1 divided by (rootx)(rootx plus 1)squared

    anyways..I got: -2/(squareroot x +1) +c

    LAST ONE =P

    6) integrate (x^2)/(1+x^6) done through two substitutions

    let u=x^3 , then let u=root3 tan theta

    In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

    arctan is tan^-1 for anyone maths illiterate reading this.

    I hope to see 6 yeses in your answers and thanks in advance =P
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by deragon999 View Post
    Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

    1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

    I got: x -5/(2x^3) -3/(x^3) -2 +c
    Incorrect:

    note that \frac {x^4 - 2x^3 + 5x + 1}{x^4} = \frac {x^4}{x^4} - \frac {2x^3}{x^4} + \frac {5x}{x^4} + \frac 1{x^4} = 1 - \frac 2x + 5x^{-3} + x^{-4}

    2) integrate: sinx/(squareroot of cosx) dx

    I got: -0.5(squareroot of cosx) +c
    Incorrect: your constant is off. check again (you can differentiate the answer to see that you are wrong)


    3) integrate (e^x)squareroot(e^x+4) dx
    I got: 0.5(squareroot e^x+4) +c
    Incorrect again: are you sure you didn't mean \int \frac {e^x}{\sqrt{e^x + 4}}~dx? your answer would still be wrong, but you would have been closer

    4) integrate xe^(x^2+1) dx

    I got: 0.5e^(x^2+1) +c
    correct

    5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
    Sorry about this 1 a bit hard to read.
    It is 1 divided by (rootx)(rootx plus 1)squared

    anyways..I got: -2/(squareroot x +1) +c
    Correct


    LAST ONE =P
    Okie dokie

    6) integrate (x^2)/(1+x^6) done through two substitutions

    let u=x^3 , then let u=root3 tan theta

    In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

    arctan is tan^-1 for anyone maths illiterate reading this.

    I hope to see 6 yeses in your answers and thanks in advance =P
    incorrect:

    just a substitution of u = x^3 will do. no second substitution is needed, and trig sub is definitely not needed
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    Just put a yes or no as to my answer =P, c is a constant, sorry about format and squareroots

    1) integrate: (x^4-2x^3+5x+1)/(x^4) dx

    I got: x -5/(2x^3) -3/(x^3) -2 +c

    2) integrate: sinx/(squareroot of cosx) dx

    I got: -0.5(squareroot of cosx) +c

    3) integrate (e^x)squareroot(e^x+4) dx
    I got: 0.5(squareroot e^x+4) +c

    4) integrate xe^(x^2+1) dx

    I got: 0.5e^(x^2+1) +c

    5) integrate 1/((squareroot x)((squareroot x) +1)^2) dx
    Sorry about this 1 a bit hard to read.
    It is 1 divided by (rootx)(rootx plus 1)squared

    anyways..I got: -2/(squareroot x +1) +c

    LAST ONE =P

    6) integrate (x^2)/(1+x^6) done through two substitutions

    let u=x^3 , then let u=root3 tan theta

    In the end i got: (squareroot 3)*arctan((x^3)/(squareroot 3)) +c

    arctan is tan^-1 for anyone maths illiterate reading this.

    I hope to see 6 yeses in your answers and thanks in advance =P
    I am feeling good tonight, I will do all of these for didactic purposes.

    1) \int\frac{x^4-2x^3+5x+1}{x^4}dx=\int\bigg[x-\frac{2}{x}+\frac{5}{x^3}+\frac{1}{x^4}\bigg]dx=x-2\ln(x)-\frac{5}{x^2}-\frac{1}{3x^3}+C

    2) \int\frac{\sin(x)}{\sqrt{\cos(x)}}dx=-\int\frac{-\sin(x)}{\sqrt{\cos(x)}}dx

    Now let \vartheta=\cos(x)\Rightarrow{d\vartheta=-\sin(x)}

    So we have

    -\int\frac{d\vartheta}{\sqrt{\vartheta}}=-2\sqrt{\vartheta}+C

    Backsubbing

    -2\sqrt{\cos(x)}+C

    3) \int{e^x\sqrt{e^x+4}dx}

    Let \varphi=e^x\Rightarrow{d\varphi=e^x}

    Giving us

    \int\sqrt{\varphi+4}d\varphi=\frac{2}{3}(\varphi+4  )^{\frac{3}{2}}+C

    backsubbing we get

    \frac{2}{3}\left(e^x+4\right)^{\frac{3}{2}}+C

    4) \int{xe^{x^2+1}dx}

    Let \psi=x^2+1\Rightarrow{d\psi=2x}
    So we have

    \frac{1}{2}\int{e^{\psi}d\psi}=\frac{1}{2}e^{\psi}  +C

    Backsubbing we get

    \frac{1}{2}e^{x^2+1}+C


    5) \int\frac{dx}{\sqrt{x}(\sqrt{x}+1)^2}

    Let \xi=\sqrt{x}\Rightarrow\xi^2=x\Rightarrow{2\xi=dx}

    So we have

    2\int\frac{\xi}{\xi(1+\xi)^2}=2\int\frac{d\xi}{(1+  \xi)^2}=\frac{-2}{1+\xi}+C

    Backsubbing we get

    \frac{-2}{1+\sqrt{x}}+C

    6) \int\frac{x^2}{1+x^6}dx

    Let \zeta=x^3\Rightarrow{d\zeta=3x^2}
    So now we have

    \frac{1}{3}\int\frac{d\zeta}{1+\zeta^2}=\frac{1}{3  }\arctan(\zeta)+C

    Now backsubbing we get

    \frac{1}{3}\arctan(x^3)+C
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  4. #4
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    Ok thanks i think i get all that, funny symbols and all =P, i just differentiated instead of integrating at some points..doh..and made a few minor errors that totally screwed me over...hmmm
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    .. funny symbols and all =P..
    You say abc I say \alpha\beta\gamma
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  6. #6
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    Well actually when the mood takes me i say a i u e o ka ki ku ke ko...

    Btw could you please check out my post just before this one..more integration gone wrong i guess...
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    Well actually when the mood takes me i say a i u e o ka ki ku ke ko...

    Btw could you please check out my post just before this one..more integration gone wrong i guess...
    What language is that?
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  8. #8
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    Jap..tho i hardly speak it since i stopped learning 2 years ago

    nihongo o wakrimasu ka
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by deragon999 View Post
    Jap..tho i hardly speak it since i stopped learning 2 years ago
    Oh...ok. Dont feel bad, I have not spoken any French this summer
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  10. #10
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    so thats a big yes to you cheking out my other post =P...all simple stuff i hope
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