# Can You Solve This Integral Without Using Integration by Parts?

• raincheck
In summary, to evaluate the integral [0,1] x^3/sqrt[x^2 + 1] by integration by parts, you can first rewrite the integral as x^3 (x^2 + 1)^-1/2 and then use the integration by parts formula, choosing u = x^2 + 1 and dv = x^3 dx. Alternatively, you can also substitute u = x^2 + 1 and dv = x^3 dx to make the process easier.
raincheck
"Evaluate the integral [0,1] x^3/sqrt[x^2 + 1] by integration by parts"

I know I have to use the integration by parts equation, but I don't know what to make u and what to make dv..

Try rewriting the integral thus;

$$\int^{1}_{0}\; \frac{x^{3}}{\sqrt{x^2+1}} \; dx = \int^{1}_{0}\; x^3 (x^2 +1)^{-\frac{1}{2}} \;dx$$

Now, to determine which term to make u and dv, think about which one will simplify your expression most when you differentiated it and set this to u.

You could start by rewriting the integrand:

$$\frac{x^3}{\sqrt{x^2+1}}=x^3 (x^2+1)^{-1/2} = [x^{-6} (x^2+1)]^{-1/2}$$

And the apply the integration by parts formula. That should make it easier.

OH okay, thank you!

Do you have to do it by parts? Much easier to substitute $$u = x^2$$

## 1. What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It is based on the product rule for differentiation and allows for the integration of certain types of functions that cannot be integrated using other methods.

## 2. When should I use integration by parts?

Integration by parts is typically used when the integrand (the function being integrated) is a product of two functions, and one of the functions becomes simpler when differentiated while the other becomes simpler when integrated.

## 3. What is the formula for integration by parts?

The formula for integration by parts is ∫u dv = uv - ∫v du, where u and v are the two functions being multiplied together and du and dv are their respective differentials.

## 4. How do I choose which function to differentiate and which to integrate?

When using integration by parts, you can choose which function to differentiate and which to integrate based on the acronym "LIATE." This stands for: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. The function closer to the beginning of this list is usually chosen for u, while the one closer to the end is chosen for dv.

## 5. Can integration by parts be used to solve definite integrals?

Yes, integration by parts can be used to solve definite integrals. In this case, the formula becomes ∫a^b u dv = [uv]a^b - ∫a^b v du, where a and b are the limits of integration. The resulting expression should be evaluated at these limits to find the final answer.

• Calculus and Beyond Homework Help
Replies
5
Views
543
• Calculus and Beyond Homework Help
Replies
1
Views
981
• Calculus and Beyond Homework Help
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
22
Views
1K
• Calculus and Beyond Homework Help
Replies
14
Views
496
• Calculus and Beyond Homework Help
Replies
2
Views
492
• Calculus and Beyond Homework Help
Replies
5
Views
850
• Calculus and Beyond Homework Help
Replies
11
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
878
• Calculus and Beyond Homework Help
Replies
9
Views
378