# Calculus 1 : How to find integral expressions

## Example Questions

### Example Question #1101 : Functions

Explanation:

To integrate this expression, you must use "u" substitution. The expression you assign to "u" is usually the expression with the higher exponent; in this case, . Since , the next step is to find du so that you can fully substitute everything.

The derivative of our u expression is

, or .

Now we can fully plug into our integral expression so we can integrate. You can rewrite the expression so it looks like this:

.

Everything in the u terms fully replaced everything in the original expression. Now, we can integrate. When we integrate , it becomes .

Since it's an indefinite expression, we must add +C at the end. The last step is to sub back in the expression that "u" represents.

.

### Example Question #52 : Integral Expressions

Explanation:

This integral requires "u" substitution. First assign an expression to u. In this case, . The next step is to find the derivative of u. Therefore, . Since there is no 2 in the original function, you must offset that by putting it on the "du" side: .

Now you can plug in everything so that the substitution is complete:

.

It might be helpful to put the  outside the integral sign and to bring the u into the numerator:

.

Now, you can integrate the u term. Raise the exponent by 1 and also put that result on the denominator: . Then multiply that by the , which gives you . Now, sub back in the expression that u was representing: . Now, evaluate that expression from . When you plug in , you get . Plugging in  yields . Now subtract those results to give you .

### Example Question #1102 : Functions

Explanation:

A good first step here is to simplify the expression. Since there's only one term on the bottom, you can make three separate terms out of the function. Therefore, rewriting it would look like:

and further simplifying it would look like this: .

Then, integrate each term. To integrate 2x, you must first raise the exponent by 1 and then put that result on the denominator, and then multiply by the coefficient 2: .

Anytime you integrate a constant, it becomes that constant multiplied by x; therefore 5x in this case. Integrating is a little trickier because of the x on the denominator. But whenever we have a single x on the denominator, we know that "ln" is involved and then we just multiply that by the coefficient: .

Also, since this is an indefinite integral, there must be a "+C" tacked on to the end.

.

### Example Question #54 : Integral Expressions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for the missing constants.

### Example Question #1103 : Functions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

### Example Question #1101 : Functions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same umber. Don't forget to add a "plus C" to account for missing constants.

### Example Question #1104 : Functions

Evaluate the following integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem we get the following for the first term:

And for the second term:

We can combine these terms and add our "C" to get the final answer:

### Example Question #1105 : Functions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

And the following for the second term:

We can combine these terms and add our "C" to get the final answer:

### Example Question #1106 : Functions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

The following for the second term:

And the following for the third term:

We can combine these terms and add our "C" to get the final answer:

### Example Question #57 : Equations

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral, use the inverse power rule.

This tells us that for each term we increase the power of x by 1 and divide by that same number. Don't forget to add a "plus C" to account for missing constants.

Applying that rule to this problem gives us the following for the first term:

The following for the second term:

And the following for the third term:

We can combine these terms and add our "C" to get the final answer: