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Thread: need help with integration by parts

  1. #1
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    need help with integration by parts

    Evaluate the following:

    • $\displaystyle \int\frac{(ln t)^2}{t}\, dt$
    • $\displaystyle \int\ln(x^2 + 1)\, dx$
    • $\displaystyle \int\sin t\, ln(cos t)\, dt$
    • $\displaystyle \int\frac{cot^-1\sqrt{z}} {\sqrt{z}}\,dz$
    • $\displaystyle \int\sqrt{x}\ ln x\,dx$
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  2. #2
    Senior Member polymerase's Avatar
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    $\displaystyle u=(ln t)^2$
    $\displaystyle dv=\dfrac{1}{t}$
    $\displaystyle du=\dfrac{2(ln t)}{t}$
    $\displaystyle v=ln |t|$

    $\displaystyle \int\frac{(ln t)^2}{t}\, dt = ln|t|(ln t)^2-\int\frac{2(ln t)^2}{t}\, dt$
    $\displaystyle 3\int\frac{(ln t)^2}{t}\, dt = {ln|t|(ln t)^2}$
    $\displaystyle \int\frac{(ln t)^2}{t}\, dt = \dfrac{(ln x)^3}{3}$

    get the idea?
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  3. #3
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    Quote Originally Posted by cazimi View Post
    Evaluate the following:
    $\displaystyle \int\ln(x^2 + 1)\, dx$
    Define $\displaystyle f(x)=x\ln(x^2+1)\implies f'(x)=\ln(x^2+1)+\frac{2x^2}{x^2+1}.$

    Now $\displaystyle \frac{{2x^2 }}
    {{x^2 + 1}} = 2\left( {1 - \frac{1}
    {{x^2 + 1}}} \right).$

    Integrate $\displaystyle f'(x),$

    $\displaystyle f(x)+k=\int\ln(x^2+1)\,dx+2(x-\arctan x).$

    And finally $\displaystyle \int\ln(x^2+1)\,dx=x\ln(x^2+1)-2(x-\arctan x)+k.$
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  4. #4
    Senior Member polymerase's Avatar
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    Let $\displaystyle u=ln(x^2+1)$ and $\displaystyle v = x$

    Thus,
    $\displaystyle du=\dfrac{2x}{x^2+1}dx$ and $\displaystyle dv = dx$

    $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-\int\dfrac{2x^2}{x^2+1}\, dx$
    $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2\int\dfrac{x^2+1-1}{x^2+1}\, dx$
    $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2\int1-\dfrac{1}{1+x^2}\, dx$
    $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2(x-arctan\:x)$
    Therefore $\displaystyle \int\ln(x^2 + 1)\, dx = x\:ln(x^2+1)-2x+arctan\:x+C$
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  5. #5
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    The first one does not require integration by parts, just define $\displaystyle u=\ln t$ and the rest follows.
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  6. #6
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    Quote Originally Posted by cazimi View Post
    $\displaystyle \int\sqrt{x}\ ln x\,dx$
    Consider the function $\displaystyle f(x)=x^{3/2}\ln x\implies f'(x)=\frac32\sqrt x\ln x+\sqrt x.$

    Now apply some of make-up,

    $\displaystyle \frac23f'(x)=\sqrt x\ln x+\frac23\sqrt x.$

    Integrate, $\displaystyle \frac23f(x)+k=\int\sqrt x\ln x\,dx+\frac49x^{3/2}.$

    Back substitute

    $\displaystyle \int\sqrt x\ln x\,dx=\frac23x^{3/2}\ln x-\frac49x^{3/2}+k.$
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  7. #7
    Senior Member polymerase's Avatar
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    Let $\displaystyle u=ln(cos\:x)$ and $\displaystyle dv=sin\:dx$
    Thus, $\displaystyle du=\dfrac{sin\:x}{cos\:x}dx$ and $\displaystyle v=-cos\:x$

    $\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:xln(cos\:x)-\int\dfrac{-sin\:x\:cos\:x}{cos\:x}dx$
    $\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:xln(cos\:x)+\int\sin\:x\:dx$

    Therefore $\displaystyle \int\sin x\, ln(cos x)\, dx = -cos\:x[ln(cos\:x)+1] + C$
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  8. #8
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    Thank you!
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  9. #9
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    integration by parts


    find the following genreal anti-derivatives using integration by parts formula


    \int\frac{(x+ 2}{e^3x}\, dx


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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Sundevils View Post
    find the following genreal anti-derivatives using integration by parts formula


    \int\frac{(x+ 2}{e^3x}\, dx


    Please do not double post. See rule #1 here.

    -Dan
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