To solve this problem we need to use u-substitution. The key to knowing that is by noticing that we have both an  and an  term, and that hypothetically if we could take the derivate of the  term it could cancel out the  term. Let's take a closer look.

Let's choose our  in this problem to be . We would need to calculate the  term since we're switching from the x variable to the u variable.

Let's take a look back at the orginal problem and begin to make substitutions.

Notice here that the x's will cancel out, leaving us with an integral with entirely the u variable.

This is a nice simple integral that results in

. Now all we need to do is replace that u with the original variable.

The problem  is a U-substitution question. The term  might not be easily seen, but the  term must equal to .

Factor the denominator by taking  as the common factor.

Rewrite the integral.

Resubstitute .

Although the integral may look tough, we see a possible relationship between the polynomial of the exponential and the polynomial in front of the exponential.

Let's make our 

Now let's see the original integral to make the substitutions.

.

At first sight it may seem that nothing cancels out, but at a closer look we can see that the  term can actually be simplified to , which can now cancel with the denominator.

 is our new integral, which simply leads to

. Now substitute u for what we had earlier.

To evaluate , use U-substitution.

Let , which also means .  Take the derivative and find .

Rewrite the integral in terms of  and , and separate into two integrals.

Evaluate the two integrals.

Re-substitute .

Pull out the common factor .

In order to solve this, we must use -substitution.

Because , we should let  so the  can cancel out.

We can now change our integral to .

We know that , so , which means .

We can substitue that in for  in the integral to get .

The  can cancel to get .

The limits of the integral have been left off because the integral is now with respect to , so the limits have changed. This integral can now be solved using the power rule  

which will give you .

Now we can substitute  back in for  to get . We can bring back our limits now and evaluate because we're in terms of  again, which is what our original limits were with respect to.

TIP: whenever doing -substitution, don't use the limits until you're in terms of the original variable.

Here, we can use u-substitution. We'll set  and we'll factor out the  outside the integral.

Now let's calculate 

And solve for dx.

.

Plugging these values into the integral we now get

.

We now see that the 's cancel out and we're left with an integral entirely with u.

.

We just need to replace u with its original value, doing so results in our final solution.

To solve the following integral, we must make a substitution to create the following general form:

We make the following subsitution:

The derivative was found using the following rule:

The integral now looks like this:

Notice that it is in the same form as the integral we want.

Now use the form from above to integrate:

To finish the problem, replace u with 2x:

.

We can simplify

by first doing a substitution, with , which gives us , which means that . So the integral becomes

The integral can be solved using two integration by parts, which will give us 

So now we just plg in  into  and  into  to get

Pull out the constant out in front of the integral.

Use U-substitution to solve.

The integral can be solved with a clever substitution:

The derivative was found using the following rules:

,

Then, when you rewrite the integral in terms of u, you find that you get:

The integration was performed using the following rule:

Finally, replace the u with our original term.