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Basic standard integral

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
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)u=g(x) and it should drop out. That integrand is in the form of the derivative of a composite function.
(edited 13 years ago)
Reply 2
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
f(x) dx=F(x)+c\displaystyle\int f(x)\ dx=F(x)+c so that
[F(x)+C]=f(x)[F(x)+C]'=f(x)
from this
f(x) dx=f(x)+c\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)g(x)[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
f(g(x))g(x) dx=f(g(x))+C\int f'(g(x)) \cdot g'(x)\ dx=f(g(x))+C
Generally
h[u(x)]u(x) dx=H[u(x)]+C\displaystyle\int h[u(x)]\cdot u'(x)\ dx=H[u(x)]+C where H is the primitive function of h.
For example:
tanx dx=(sinx)1cosx dx=lncosx+C\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.
1xlnx dx=1x1lnx dx=lnlnx+C\displaystyle\int \frac{1}{xlnx}\ dx=\int \frac{1}{x}\cdot\frac{1}{lnx}\ dx=ln|lnx|+C
Unparseable latex formula:

\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

(edited 13 years ago)
Try differentiating f(g(x)). Use a substitution u = g(x).

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