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Basic standard integral

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    Basically we have been given a sheet with all the standard integrals we should know.

    I don't however understand this one:

    Integral f'(g(x))g'(x)dx = f(g(x)) + c

    Would someone using latex just give an example using a simple equation as then I should see it much clearer!

    Thanks
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    (Original post by TimothyTankT)
    Basically we have been given a sheet with all the standard integrals we should know.

    I don't however understand this one:

    Integral f'(g(x))g'(x)dx = f(g(x)) + c

    Would someone using latex just give an example using a simple equation as then I should see it much clearer!

    Thanks
    Use a substitution of u=g(x) and it should drop out. That integrand is in the form of the derivative of a composite function.
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    (Original post by TimothyTankT)
    Basically we have been given a sheet with all the standard integrals we should know.

    I don't however understand this one:

    Integral f'(g(x))g'(x)dx = f(g(x)) + c

    Would someone using latex just give an example using a simple equation as then I should see it much clearer!

    Thanks
    It is the rule for the integration of compodite functions and the base of this notation
    is the definition of undetermined integral namely
    \displaystyle\int f(x)\ dx=F(x)+c so that
    [F(x)+C]'=f(x)
    from this
    \displaystyle \int f'(x)\ dx=f(x)+c
    If function f in composite, for example f(g(x)), then from the chain rule
    [f(g(x))]'=f'(g(x)\cdot g'(x) where f' is derivative of outer function
    substituting g(x) in x multiplied by the derivative of inner function, g'(x).
    So integrating this
    \int f'(g(x)) \cdot g'(x)\ dx=f(g(x))+C
    Generally
    \displaystyle\int h[u(x)]\cdot u'(x)\ dx=H[u(x)]+C where H is the primitive function of h.
    For example:
    \displaystyle\int -tanx\ dx=\int (-sinx)\frac{1}{cosx}\ dx=ln|cosx|+C
    because the primitive function of 1/x is ln|x| and the derivative of cosx is -sinx.
    F.e 2.
    \displaystyle\int \frac{1}{xlnx}\ dx=\int \frac{1}{x}\cdot\frac{1}{lnx}\ dx=ln|lnx|+C
    \displaystyle\int \frac{x}{1+x^4}\ dx=\frac{1}{2}\int 2x\frac{1}{1+(x^2)^2}\dx=\frac{1  }{2}arctan(x^2)+C
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    Try differentiating f(g(x)). Use a substitution u = g(x).

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Updated: October 28, 2010
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