Recall the Maclaurin series formula:

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is  as its Maclaurin series is.

Using the formula of a binomial series centered at 0:

 ,

where we replace  with  and , we get:

  for the first 3 terms.

Then, we find the terms where,

By the ratio test, the series  converges absolutely: 

Using the root test

and

. T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

To get , we need to find the antiderivative of  by integrating the third degree polynomial term by term.

We only want up to a third degree polynomial, so we can disregard the fourth order term:

Since , substitute  for the final .

The Taylor series about x=a of any function is given by the following:

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

The derivatives were found using the following rule:

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

which when simplified is equal to

.

The general formula for the Taylor series about x=a for a function is

First, we must find the zeroth and first derivative of the function.

The zeroth derivative of a function is just the function itself, so we only have to find the first derivative:

The derivative was found using the following rule:

Now, write the first two terms of the sequence (n=0 and n=1):

The general formula for the Taylor series of a given function about x=a is

.

We were asked to find the first three terms, which correspond to n=0, 1, and 2. So first, we need to find the zeroth, first, and second derivative of the given function. The zeroth derivative is just the function itself.

The derivatives were found using the following rules:

, 

Now use the above formula to write out the first three terms:

Simplified, this becomes

This problem requires U-Substitution.  Let  and find .

Notice that the numerator in  has common factor of 2, 3, or 6.  The numerator can be factored so that the  term can be a substitute. Factor the numerator using 3 as the common factor.

Substitute  and  terms, integrate, and resubstitute the  term.

To calculate this integral, we could expand that whole binomial, but it would be very time consuming and a bit of a pain. Instead, let's use u substitution:

Given this:

We can say that 

Then, plug it back into our original expression

Evaluate this integral to get

Then, replace u with what we substituted it for to get our final answer. Note because this is an indefinite integral, we need a plus c in it.