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Thread: indefinite integrals (lots of them)

  1. #1
    Super Member Random Variable's Avatar
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    indefinite integrals (lots of them)

    These were posted on another forum.


    1) $\displaystyle \displaystyle \int \frac{dx}{\sqrt{5x+3}+\sqrt{5x-2}}$


    2) $\displaystyle \displaystyle \int \frac{dx}{\sqrt{\cos^{3}(x) \sin^{5}(x)}}$


    3) $\displaystyle \displaystyle \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^{2}+1}} \ dx $


    4) $\displaystyle \displaystyle \int \frac{dx}{\sqrt{x+x\sqrt{x}}} $


    5) $\displaystyle \displaystyle \int e^{2x} \ \frac{2x-1}{4x^{2}} \ dx $


    6) $\displaystyle \displaystyle \int \frac{dx}{x(1+\ln x) \sqrt{\ln x(2+\ln x)}} $


    7) $\displaystyle \displaystyle \int \frac{x^{4}+1}{x^{2}\sqrt{x^4-1}} \ dx $


    8) $\displaystyle \displaystyle \int \ln (\sqrt{x} + \sqrt{1+x}) \ dx $


    9) $\displaystyle \displaystyle \int \frac{dx}{x- \sqrt{1-x^{2}}} \ dx $


    10) $\displaystyle \displaystyle \int \frac{dx}{\sin x + \sqrt{1-\sin 2x} } $


    11) $\displaystyle \displaystyle \int \frac{dx}{(2x-1)\sqrt{x^{2}-x}} $


    12) $\displaystyle \displaystyle \int \frac{2e^{2x}-e^{x}}{\sqrt{3e^{2x}-6e^{x}-1}} \ dx $
    Last edited by Random Variable; Feb 25th 2011 at 07:11 AM.
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  2. #2
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    Quote Originally Posted by Random Variable View Post
    These were posted on another forum.


    $\displaystyle \displaystyle \int \frac{dx}{\sqrt{5x+3}+\sqrt{5x-2}}$


    $\displaystyle \displaystyle \int \frac{dx}{\sqrt{\cos^{3}(x) \sin^{5}(x)}}$


    $\displaystyle \displaystyle \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^{2}+1}} \ dx $


    $\displaystyle \displaystyle \int \frac{dx}{\sqrt{x+x\sqrt{x}}} $


    $\displaystyle \displaystyle \int e^{2x} \ \frac{2x-1}{4x^{2}} \ dx $


    $\displaystyle \displaystyle \int \frac{dx}{x(1+\ln x) \sqrt{\ln x(2+\ln x)}} $


    $\displaystyle \displaystyle \int \frac{x^{4}+1}{x^{2}\sqrt{x^4-1}} \ dx $


    $\displaystyle \displaystyle \int \ln (\sqrt{x} + \sqrt{1+x}) \ dx $


    $\displaystyle \displaystyle \int \frac{dx}{x- \sqrt{1-x^{2}}} \ dx $


    $\displaystyle \displaystyle \int \frac{dx}{\sin x + \sqrt{1-\sin 2x} } $


    $\displaystyle \displaystyle \int \frac{dx}{(2x-1)\sqrt{x^{2}-x}} $


    $\displaystyle \displaystyle \int \frac{2e^{2x}-e^{x}}{\sqrt{3e^{2x}-6e^{x}-1}} \ dx $
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    Thanks for posting. Could you number them please, if it isn't much of a trouble?

    $\displaystyle \displaystyle \begin{aligned} 1. \int \frac{dx}{\sqrt{5x+3}+\sqrt{5x-2}} & = \int \frac{\sqrt{5x+3}-\sqrt{5x-2}}{\left(\sqrt{5x+3}+\sqrt{5x-2}\right)\left(\sqrt{5x+3}-\sqrt{5x-2}\right)}\;{dx} \\ & = \int \frac{\sqrt{5x+3}-\sqrt{5x-2}}{(5x+3)-(5x-2)}\;{dx} = \frac{1}{5}\int \sqrt{5x+3}-\sqrt{5x-2}}\;{dx} \\ & = \frac{2}{5\times 5 \times 3}\left(5x+3\right)^{\frac{3}{2}}-\frac{2}{5\times 5 \times 3}\left(5x-2\right)^{\frac{3}{2}}+k \\ & = \frac{2}{75}\sqrt{(5x+3)^3}-\frac{2}{75}\sqrt{(5x-2)^3}+k. \end{aligned}$

    Quote Originally Posted by Prove It View Post
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    $\displaystyle \displaystyle \int{\frac{dx}{x - \sqrt{1 - x^2}}}$.

    Make the substitution $\displaystyle \displaystyle x = \sin{\theta} \implies dx = \cos{\theta}\,d\theta$ and the integral becomes

    $\displaystyle \displaystyle \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta} - \sqrt{1 - \sin^2{\theta}}}}$

    $\displaystyle \displaystyle = \int{\frac{\cos{\theta}\,d\theta}{\sin{\theta} - \cos{\theta}}}$

    $\displaystyle \displaystyle = \int{\frac{\cos{\theta}(\sin{\theta} + \cos{\theta})\,d\theta}{\sin^2{\theta} - \cos^2{\theta}}}$

    $\displaystyle \displaystyle = \int{\frac{\cos{\theta}\sin{\theta} + \cos^2{\theta}}{-(\cos^2{\theta} - \sin^2{\theta})}}$

    $\displaystyle \displaystyle = \int{\frac{\frac{1}{2}\sin{2\theta} + \frac{1}{2} + \frac{1}{2}\cos{2\theta}}{-\cos{2\theta}}}$

    $\displaystyle \displaystyle = -\frac{1}{2}\int{\frac{\sin{2\theta} + 1 + \cos{2\theta}}{\cos{2\theta}}}$

    $\displaystyle \displaystyle = -\frac{1}{2}\int{\tan{2\theta}\,d\theta}-\frac{1}{2}\int{\sec{2\theta}\,d\theta} - \frac{1}{2}\int{d\theta}$

    $\displaystyle \displaystyle = \frac{1}{4}\ln{|\cos{2\theta}|} - \frac{1}{4}\ln{|\tan{2\theta} + \sec{2\theta}|}- \frac{\theta}{2} + C$

    $\displaystyle \displaystyle = \frac{1}{4}\ln{\left|\frac{\cos{2\theta}}{\tan{2\t heta} + \sec{2\theta}}\right|} - \frac{\theta}{2} + C$

    $\displaystyle \displaystyle = \frac{1}{4}\ln{\left|-2\sin{x}\sqrt{1 - \sin^2{x}}- 1\right|} - \frac{\theta}{2} + C$

    $\displaystyle \displaystyle = \frac{1}{4}\ln{\left|-2x\sqrt{1 - x^2} - 1\right|} - \frac{1}{2}\arcsin{x} + C$.
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  5. #5
    Super Member Random Variable's Avatar
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    Surely an MHF Expert knows to only post 2 questions per thread...
    It's just for fun.
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    $\displaystyle \displaystyle \int{\frac{2e^{2x} - e^x}{\sqrt{3e^{2x} - 6e^x - 1}}\,dx} = \int{\frac{(2e^x - 1)e^x}{\sqrt{3\left(e^x\right)^2 - 6e^x - 1}}\,dx}$

    Let $\displaystyle \displaystyle u = e^x \implies du = e^x\,dx$ and the integral becomes

    $\displaystyle \displaystyle \int{\frac{2u - 1}{\sqrt{3u^2 - 6u - 1}}\,du}$

    $\displaystyle \displaystyle = \int{\frac{2u-1}{\sqrt{3\left(u^2 - 2u - \frac{1}{3}\right)}}\,du}$

    $\displaystyle \displaystyle = \int{\frac{2u-1}{\sqrt{3\left[u^2 - 2u + (-1)^2 - (-1)^2 - \frac{1}{3}\right]}}\,du}$

    $\displaystyle \displaystyle = \int{\frac{2u - 1}{\sqrt{3\left[(u - 1)^2 - \frac{4}{3}\right]}}\,du}$

    $\displaystyle \displaystyle = \int{\frac{2u - 1}{\sqrt{3(u - 1)^2 - 4}}\,du}$.

    Now let $\displaystyle \displaystyle u - 1 = \frac{2}{\sqrt{3}}\cosh{t} \implies du = \frac{2}{\sqrt{3}}\sinh{t}\,dt$ and the integral becomes

    $\displaystyle \displaystyle = \int{\left(\frac{\frac{4}{\sqrt{3}}\cosh{t} + 1}{2\sinh{t}}\right)\frac{2}{\sqrt{3}}\sinh{t}\,dt }$

    $\displaystyle \displaystyle = \int{\frac{1}{\sqrt{3}}\left(\frac{4}{\sqrt{3}}\co sh{t} + 1\right)\,dt}$

    $\displaystyle \displaystyle = \frac{1}{\sqrt{3}}\left(\frac{4}{\sqrt{3}}\sinh{t} + t\right) + C$

    $\displaystyle \displaystyle = \frac{4}{3}\sinh{t} + \frac{1}{\sqrt{3}}t + C$

    $\displaystyle \displaystyle = \frac{4}{3}\sqrt{\cosh^2{t} - 1} + \frac{\sqrt{3}}{3}t + C$

    $\displaystyle \displaystyle = \frac{4}{3}\sqrt{\left[\frac{\sqrt{3}}{2}(u - 1)\right]^2 - 1} + \frac{\sqrt{3}}{3}\cosh^{-1}{\left[\frac{\sqrt{3}}{2}(u - 1)\right]} + C$

    $\displaystyle \displaystyle = \frac{4}{3}\sqrt{\frac{3}{4}u^2 - \frac{3}{2}u - \frac{1}{4}} + \frac{\sqrt{3}}{3}\cosh^{-1}{\left[\frac{\sqrt{3}}{2}(u - 1)\right]} + C$

    $\displaystyle \displaystyle = \frac{4}{3}\sqrt{\frac{3}{4}e^{2x} - \frac{3}{2}e^x - \frac{1}{4}} + \frac{\sqrt{3}}{3}\cosh^{-1}{\left[\frac{\sqrt{3}}{2}(e^x - 1)\right]} + C$.

    This can probably be simplified more using logarithmic equivalents, but I think this is sufficient...
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  7. #7
    Super Member Random Variable's Avatar
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    3) $\displaystyle \displaystyle \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^{2}+1}} \ dx $


    let $\displaystyle \displaystyle u = \sqrt{x+\sqrt{x^{2}+1}} $

    $\displaystyle \displaystyle du = \frac{1}{2} \frac{1}{\sqrt{x+\sqrt{x^{2}+1}}} \Big( 1 + \frac{x}{\sqrt{x^{2}+1}} \Big) \ dx $

    $\displaystyle \displaystyle = \frac{1}{2u} \frac{u^{2}}{\sqrt{x^{2}+1}} \ dx = \frac{u}{2} \frac{1}{\sqrt{x^{2}+1}} \ dx$


    so $\displaystyle \displaystyle \int \frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^{2}+1}} \ dx = 2 \int \frac{ u \ du}{u} = 2 u + C $

    $\displaystyle \displaystyle = 2 \sqrt{x + \sqrt{x^{2}+1}}} + C $
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    4)

    $\displaystyle \displaystyle \int \frac{dx}{\sqrt{x+x\sqrt{x}}}=\int \frac{dx}{\sqrt{x}\sqrt{1+\sqrt{x}}}$

    $\displaystyle \displaystyle u=1+\sqrt{x}$ gives

    $\displaystyle \displaystyle 2\int u^{-\frac{1}{2}}du=4\sqrt{u}=4\sqrt{1+\sqrt{x}}+C$
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    6)

    let $\displaystyle \displaystyle t^2=(\ln(x)+1)^2-1 \implies tdt=\frac{\ln(x)+1}{x}dx \iff \frac{tdt}{\sqrt{t^2+1}}=\frac{dx}{x}$

    This gives

    $\displaystyle \displaystyle \int \frac{dx}{x(1+\ln(x))\sqrt{(\ln(x)+1)^2-1}}=\int \frac{dt}{t^2+1}=\tan^{-1}(\sqrt{(\ln(x)+1)^2-1 })+C$
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  10. #10
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    $\displaystyle \displaystyle \int \frac{x^{4}+1}{x^{2} \sqrt{x^{4}-1}} \ dx $


    $\displaystyle \displaystyle = \int \frac{x^{2} \big( x^{2} + \frac{1}{x^{2}} \big)}{x^{2}\sqrt{x^{2}\big(x^{2}-\frac{1}{x^{2}}\big)}} \ dx $


    $\displaystyle \displaystyle = \int \frac{x^{2} + \frac{1}{x^{2}}}{x\sqrt{x^{2}-\frac{1}{x^{2}}}} \ dx $


    let $\displaystyle x = e^{u} $


    $\displaystyle \displaystyle = \int \frac{e^{2u} + e^{-2u}}{\sqrt{e^{2u}-e^{-2u}}} \ du $


    let $\displaystyle \displaystyle w = e^{2u}-e^{-2u} $


    $\displaystyle \displaystyle = \frac{1}{2} \int \frac{dw}{\sqrt{w}} $


    $\displaystyle \displaystyle = \sqrt{w} + C = \sqrt{e^{2u}-e^{-2u}} + C $


    $\displaystyle \displaystyle = \sqrt{x^{2}- \frac{1}{x^{2}}} + C = \frac{\sqrt{x^{4}-1}}{x} + C $
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    5)

    $\displaystyle \displaystyle \int (2x-1)e^{2x}\cdot \frac{1}{4x^2}dx$

    Integrate by parts with

    $\displaystyle u=(2x-1)e^{2x} \implies du =4xe^{2x}dx$ and
    $\displaystyle dv=\frac{1}{4x^2}dx \implies v=-\frac{1}{4x}$

    This gives

    $\displaystyle \displaystyle \left( -\frac{1}{4x}\right)(2x-1)e^{2x}-\int\left( -\frac{1}{4x}\right) \left( 4xe^{2x}dx\right) $

    $\displaystyle \displaystyle -\frac{e^{2x}}{2}+\frac{e^{2x}}{4x}+\int e^{2x}dx=\frac{e^{2x}}{4x}$
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    An altenative solution for #5:

    $\displaystyle \displaystyle 5. \int \frac{e^{2x}(2x-1)}{4x^2}}\;{dx} = \int \frac{2xe^{2x}-e^{2x}}{(2x)^2}\;{dx} = \frac{1}{2}\int\frac{(e^{2x})'(2x)-(e^{2x})(2x)'}{(2x)^2}\;{dx} = \frac{1}{2}\left(\frac{e^{2x}}{2x}\right)+k.$
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    $\displaystyle ~ \displaystyle \begin{aligned} 7. ~ \int \frac{x^{4}+1}{x^{2} \sqrt{x^{4}-1}} \ dx & = \int \frac{x+\frac{1}{x^3}}{\sqrt{x^2-\frac{1}{x^2}}}\;{dx} = \int\frac{2x+\frac{2}{x^3}}{2\sqrt{x^2-\frac{1}{x^2}}}\;{dx} \\& = \int\frac{\left(x^2-\frac{1}{x^2}\right)'}{2\sqrt{x^2-\frac{1}{x^2}}}\;{dx} = \sqrt{x^2-\frac{1}{x^2}}+k \\& \because \displaystyle \int\frac{f'(x)}{2\sqrt{f(x)}} = \sqrt{f(x)}+k. \end{aligned} $
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