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To recap: to integrate is to find the antiderivative of a function, i.e. to find a function that, taken the derivative of it, gives our original function back. This means that we can find integration rules just by using the derivative rules in the opposite direction.

One if the derivative rules is the chain rule:

For example

Working backwards that means that

The constant *C* since the derivative of an added constant is 0.

It might be hard to see how to do this, but happily there is method to do this. This trick can be useful when the integrand (the equation we want to integrate) is a factor of two expressions.

To start with we have to look at if the derivative of a function or inner function is equal to the other factor, i.e. if we have

or

We call the function *f *(…) the outer function and *g*(*x*) the inner function.

(The first case is just a special case of the second. The function *f* is just the identity function that does nothing.)

In our example above we have

As we will see we can be of by a constant factor, and the trick we will use will work.

**The substitution**

Let us do the substitution

Now let us take the derivative of this with respect to *x*. We get

Let us multiply this by *dx*. Can we do that? In this case we will see that it will work. The proof is in the pudding as one say.

Now, this will give us

Finally we replace back the expression for *u*:

**In practice**

Ok, let us try this on our original example.

Here we can see that the derivative of *x*^{3}+2 is 3*x*^{2}, so we are only off with a constant factor, but that is Ok as we will see.

In the above I indicate the whole process of the substitution with a vertical line, then I continue with an “=” sign, indicating that we simply continue with the integral again. In the part with the vertical line our goal is to get to something we could easily substitute away – in this case *x*^{2}*dx*.

I would usually go directly from *u*=*x*^{3}+2 to *du*=3*x*^{2}*dx* (I would differentiate).

**A few examples**

Here it might be a bit harder to see what the outer function is, but it is 1 divided by *something* ,

i.e. the reciprocal function that has the integral ln(*of the something*).

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