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Math Help - Integration please help

  1. #1
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    Integration please help

    I need solution step by step for these integrals
    I have Maple but don't have options show me step by step
    Could you help
    Thanks

    1. int(3*t/(4*t^2+4*t-11)^1/2,t)
    2. int((x^3-x^2+x+1)/(x^2+1)*(x-1)^2,x);
    3. int(exp(((x)^1/2)),x);
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  2. #2
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    Here's a link to the first one. Was that you?.

    http://math2.org/mmb/thread/36961


    #2 use partial fractions: \int\left(\frac{1}{(x-1)^{2}}+\frac{x}{x^{2}+1}\right)dx


    For #3, try the subsitution u=\sqrt{x}

    u^{2}=x\;\;  2udu=dx
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  3. #3
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    Hello, simfonija!

    Here's my approach to #1 . . .


    1)\;\;\int \frac{3t}{\sqrt{4t^2 + 4t - 11}}\,dt

    Complete the square: . 4t^2 + 4t - 11\;=\;4t^2 + 4t + 1 - 11\;=\;(2t + 1)^2 - 12

    So we have: . 3\int\frac{t}{\sqrt{(2t+1)^2 - 12}}\,dt

    Let 2t+1 \,=\,2\sqrt{3}\sec\theta\quad\Rightarrow\quad t \,=\,\frac{1}{2}(2\sqrt{3}\sec\theta - 1)\quad\Rightarrow\quad dt  \,= \,\sqrt{3}\sec\theta\tan\theta\,d\theta

    and the radical is: . \sqrt{(2t+1)^2-12}\:=\:\sqrt{(2\sqrt{3}\sec\theta)^2 - 12}

    . . . =\:\sqrt{12\sec^2\theta - 12} \:=\:\sqrt{12(\sec^2\theta - 1)}  \:=\:\sqrt{12\tan^2\theta} \:=\:2\sqrt{3}\tan\theta

    Substitute: . 3\int\frac{\frac{1}{2}(2\sqrt{3}\sec\theta - 1}{2\sqrt{3}\tan\theta}\,(\sqrt{3}\sec\theta\tan \theta\,d \theta) \;= \frac{3}{4}\int\sec\theta(2\sqrt{3}\sec\theta - 1)\,d\theta

    . . = \;\frac{3}{4}\int\left(2\sqrt{3}\sec^2\theta - \sec\theta)\,d\theta \:= \:\frac{3}{4}\left[2\sqrt{3}\tan\theta - \ln\left|\sec\theta + \tan\theta\right|\right] + C


    Back-substitute: . \sec\theta \:=\:\frac{2t + 1}{2\sqrt{3}}\quad\Rightarrow\quad \tan\theta \:=\:\frac{\sqrt{4t^2 + 4t - 11}}{2\sqrt{3}}

    We have: . \frac{3}{4}\bigg[2\sqrt{3}\cdot\frac{\sqrt{4t^2+4t-11}}{2\sqrt{3}} - \ln\left|\frac{2t+1}{2\sqrt{3}} + \frac{\sqrt{4t^2+4t-11}}{2\sqrt{3}}\right|\,\bigg] + C<br />

    . . = \;\frac{3}{4}\bigg[\sqrt{4t^2+4t-11} - \ln\left|2t + 1 + \sqrt{4t^2 + 4t - 11\right|\,\bigg] + C

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  4. #4
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    Quote Originally Posted by simfonija
    3. int(exp(((x)^1/2)),x);
    You have,
    \int e^{\sqrt{x}} dx
    Express as,
    2\int \frac{\sqrt{x}}{2\sqrt{x}}e^{ \sqrt{x} } dx
    Let, u=\sqrt{x} then, \frac{du}{dx}=\frac{1}{2\sqrt{x}}
    Thus, you have,
    2\int \sqrt{x} e^{\sqrt{x}}\cdot \frac{1}{2\sqrt{x}} dx
    Becomes,
    2\int u e^u \frac{du}{dx}dx=2\int ue^udu
    Ignore the 2 and use integration by parts,
    v=u then, v'=1
    w'=e^u then, w=e^u
    Thus, by parts,
    ue^u-\int e^u du
    Thus,
    ue^u-e^u+C
    Multiply by two,
    2ue^u-2e^u+C
    Substitute,
    2\sqrt{x} e^{\sqrt{x}} -2e^{\sqrt{x}}+C
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  5. #5
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    Hello, simfonija!

    Galactus had the best idea for #3 . . .


    3)\;\;\int e^{\sqrt{x}}\,dx

    Let \sqrt{x} = y\quad\Rightarrow\quad x = y^2\quad\Rightarrow\quad dx = 2y\,dy

    Substitute: . \int e^y(2y\,dy) \;= \;2\int y\,e^y\,dy


    Integrate by parts:

    . . Let: . u = y\qquad dv = e^y\,dy

    . . Then: du = dy\quad v = e^y

    And we have: . 2\left[y\!\cdot\! e^y - \int e^u\,du\right] \;=\;2\left(y\!\cdot\! e^y - e^y\right) + C = \;2e^y(y - 1) + C


    Back-substitute: . \boxed{2e^{\sqrt{x}}(\sqrt{x} - 1) + C}

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  6. #6
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    Thank you so much to all, can somebody help with the last one, I get lost in coificients, cant find A,B... I have just 15 min to submit my assignment
    Thanks
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  7. #7
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    #2:

    Do the partial fraction thing:

    \frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{Cx+D}{x^{2}+1}

    (x^{2}+1)(x-1)A+B(x^{2}+1)+(Cx+D)(x-1)^{2} =x^{3}-x^{2}+x+1

    This leads to the following system:

    A+C=1;\\-A+B-2C+D= 1;\\A+C-2D+1;\\-A+B+D=1

    The solutions are A=0;\;\ B+1;\;\ C=1;\;\ D=0

    Break it into 2 integrals after you do the partial fractions.

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

    For the first half:

    u=x-1;\;\ u+1=x;\;\ du=dx

    \int\frac{1}{u^{2}}du=\frac{-1}{u}=\frac{-1}{x-1}

    For the 2nd half:

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

    u=x^{2}+1;\;\ du=2xdx;\;\frac{du}{2}=xdx

    \frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln(u)=\fra  c{1}{2}ln(x^{2}+1)

    So, put 'er all together:

    \frac{-1}{x-1}+\frac{1}{2}ln(x^{2}+1)
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  8. #8
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    simfonija

    Thanks again, great help.
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