Evaluating this integral requires a u-substitution. One must also notice that this integral takes the form of the derivative of :

Now we arrive at the final answer by plugging in for u.

At first glance, you notice that this integral is composed of two functions multiplied by each other. A simple "u" substitution wouldn't suffice. We can try that, and see where that will take us:

If we let , then . We then have the  term to worry about, and there isn't any manipulation that can be done with these terms to somehow turn this into an integral in terms of "u". When this simple "u" substitution doesn't work, we generally can test to see if integration by parts is a valid approach. Integration by parts is just a simple application of a formula:

When given: 

 Where "u" is a differentiable function, and "dv" is a term that can be integrated. 

The "difficult" part of applying this formula is to figure out which term will be set equal to "u", and which term will equal "dv".

Thankfully, there is a general approach that works in most cases, when dealing with these kinds of problems. 

All one needs to know is the acronymn, L.I.A.T.E. What do these letters stand for? Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This heirarchy can be followed when deciding on the "u" term. For example, in our given problem statement, we have an Algebraic Function with , and we have a Logarithmic Function with . We therefore, will set . 

So now we have:

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 will simply be the derivative of the "u" term with respect to "x". "v" will simply be the integral of the "dv" term with respect to x. We arrive at:

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Applying these terms into the formula for Integration By Parts results in:

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Evaluating this integral requires use of a substitution. Generally with an integral involving exponential functions, we should try setting u equal to the function that e is being raised to, like so:

Making these substitutions with the given integral we arrive at:

This is one of the answer choices.

Evaluating this integral requires knowledge of the inverse trigonometric functions, particuarly .

Now if we make a substitution:

The integral turns into:

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A trigonometric transformation must first be utilized before solving the integral.

For:

Using this formula, the integral turns into:

Now that we have two separate integrals which can be easily integrated like so:

Now that we've made the substitutions, the integrals become:

Now substituting back in for u and v, we arrive at the final answer:

Evaluation of this integral requires making a substitution.

This is the typical substitution we would make. Notice that doing this substitution only affects the term under the "square-root" symbol, and we still have the lone x term to worry about. Therefore we do the following:

We know that , so it is also true that .

Now arrange the integral using these substitutions:

Solving this integral is relatively simple:

We now substitute back in for u:

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For:

When asked to write out the form of Partial Fraction Decomposition, we can ignore the numerator p(x) and simply look at the denominator q(x) of the given function. In our case, the denominator:

We can further break this up into two separate parts:

 where 

 falls into the Linear Factor Rule:

"For each factor of the form , the partial fraction decomposition contains the following sum of m partial fractions:

where  are constants to be determined."

- Howard Anton "Calculus"

For our case,  can be put into the following partial fraction decomposition form:

 falls into the Quadratic Factor Rule:

"For each factor of the form , the partial fraction decomposition contain the following sum of m partial fractions:

where  are constants to be determined."

- Howard Anton "Calculus"

For our case,  can be put into the following partial fraction decomposition form (Note A and B have already been taken, so we will start with C)

Putting it all together our final answer is:

At first glance we notice that this is an improper integral of Type I, because it has an infinite interval. For cases like this, the integral is solved like so:

Applying this to our problem statement:

If we ignore the bounds and were to solve this integral, we would use a simple u-substitution like so:

Since there are bounds to this integral we can ignore the "+C" term.

Evaluating this results in:

This is one of the answer choices.

Simplify the integral using the trig identity as follows,

Rewrite the integral as follows,

Further simplify the integral using the trig identity as follows,

Rewrite the integral as

Pull the constants outside of the integral

Integrate each individual term using integration of a constant and integration of a trig function.

Resulting in the expression

Solve for the definite integral using the boundaries given.

Resulting in the answer

Use partial fraction decomposition on the expression in the integral as follows,

Set  then,

Set  then, 

Set  then,

Therefore we get,

Rewrite the integral as follows,

Now integrate each term individually using the follows integration rules:

When ,

Resulting in our final answer,