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Thread: integral Tania

  1. #1
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    Smile integral Tania

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

    somebody help ....
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Use the trig substitution x =tan(t)

    you will end up with the integral of cos^4(t) which can use a reduction formula on or

    write cos^4(t) =cos^2(t) - cos^2(t)sin^2(t) = cos^2(t) - 1/4 sin^2(2t)


    cos^2(t) - 1/4 sin^2(2t) = (1+cos(2t)/2 -1/8(1- cos(4t))
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    $\displaystyle \int{\frac{x^{2}}{\left( x^{2}+1 \right)^{2}}\,dx}=\int{x\left( -\frac{1}{2\left( x^{2}+1 \right)} \right)'\,dx}=-\frac{x}{2\left( x^{2}+1 \right)}+\frac{1}{2}\arctan (x)+k_1.$ Since $\displaystyle \frac{1}{x^{2}\left( x^{2}+1 \right)^{2}}=\frac{1}{x^{2}}-\frac{2}{x^{2}+1}+\frac{x^{2}}{\left( x^{2}+1 \right)^{2}},$ then:

    $\displaystyle \begin{aligned}
    \int{\frac{dx}{\left( 1+x^{2} \right)^{3}}}&=\int{\frac{x}{x\left( 1+x^{2} \right)^{3}}\,dx}\\
    &=\int{\frac{1}{x}\left( -\frac{1}{4\left( x^{2}+1 \right)^{2}} \right)'\,dx}\\
    &=-\frac{1}{4x\left(x^{2}+1 \right)^{2}}-\frac{1}{4}\int{\frac{dx}{x^{2}\left( x^{2}+1 \right)^{2}}}\\
    &=-\frac{1}{4x\left(x^{2}+1\right)^{2}}+\frac{1}{4}\l eft(\frac{1}{x}+\frac{3}{2}\arctan (x)+\frac{x}{2\left(x^{2}+1\right)}\right)+k.
    \end{aligned}$
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