# Solving Substitution Integrals: Guide and Example Problems

• 3.141592654
In summary: If it does, you've found the error. Otherwise, you might want to ask for help from someone who has done this before.In summary, there may be a typo in your first integral, but the rest of your work appears to be correct. You may want to double check your work or ask for help from someone who has done this type of problem before.
3.141592654

## Homework Statement

$$\int(x^5\sqrt{x^2+4})dx$$

The answer is given as: $$=105(x^2+4)^\frac{3}{2}(15x^4-48x^2+128)+C$$

## The Attempt at a Solution

$$u=\sqrt{x^2+4}$$

$$u^2=x^2+4$$

$$2udu=2xdx$$

$$udu=xdx$$

$$u^2-4=x^2$$

$$\int(x^5\sqrt{x^2+4})dx = \int(u^2+4)^2u^2du = \int(u^4-8u^2+16)u^2du$$

$$=\int(u^6-8u^4+16u^2)du = \int(u^6)du-8\int(u^4)du+16\int(u^2)du$$

$$=\frac{u^7}{7}-8\frac{u^5}{5}+16\frac{u^3}{3}+C$$

$$=\frac{(x^2+4)^\frac{7}{2}}{7}-8\frac{(x^2+4)^\frac{5}{2}}{5}+16\frac{(x^2+4)^\frac{3}{2}}{3}+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{(x^2+4)^\frac{4}{2}}{7}-8\frac{(x^2+4)^\frac{2}{2}}{5}+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}(x^2+4)^2-\frac{8}{5}(x^2+4)+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}(x^4-8x^2+16)-\frac{8}{5}(x^2+4)+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}[\frac{1}{7}x^4-\frac{8}{7}x^2+\frac{16}{7}+\frac{8}{5}x^2+\frac{32}{5}+\frac{16}{3}]+C$$

$$=(x^2+4)^\frac{3}{2}(\frac{1}{7}x^4+\frac{16}{35}x^2+\frac{1248}{105})+C$$

I'm obviously off by a bit from the stated answer, but I can't find where I may have went wrong.

I don't see anything obviously wrong, other than a typo or two that didn't affect your calculuations.
$$\int(x^5\sqrt{x^2+4})dx = \int(u^2+4)^2u^2du = \int(u^4-8u^2+16)u^2du$$
The first integral in u should have (u^2 - 4), not (u^2 + 4), but the following expression is correct.

In the next line, you have your antiderivative. As a sanity check, take its derivative, and if you get the original integrand, your work is good.
$$=\frac{(x^2+4)^\frac{7}{2}}{7}-8\frac{(x^2+4)^\frac{5}{2}}{5}+16\frac{(x^2+4)^\fr ac{3}{2}}{3}+C$$

As a final check, if the derivative of your last expression is equal to the integrand, then your answer works. If that's the case, you might want to differentiate the answer you are given and see if it works, also.

## 1. What is the substitution method in integration?

The substitution method, also known as the u-substitution method, is a technique used in integration to simplify the integrand and make it easier to solve. It involves substituting a variable with a simpler expression, usually denoted as u, and then solving the integral in terms of u before converting it back to the original variable.

## 2. When should I use the substitution method in integration?

The substitution method is useful when the integrand contains a complicated function or expression that can be simplified by substitution. It is also helpful when the integrand contains functions that are the derivatives of each other, making it easier to integrate.

## 3. How do I choose the right substitution for an integral?

To choose the right substitution, look for a part of the integrand that can be represented as a single variable. This could be a function, a term, or a combination of both. Then, let u be this variable and substitute it into the integral.

## 4. What is the difference between substitution and integration by parts?

Substitution and integration by parts are two different techniques used to solve integrals. In substitution, a variable is substituted to simplify the integrand, while in integration by parts, the integrand is split into two parts and integrated separately.

## 5. Are there any tips for solving difficult substitution integrals?

One tip for solving difficult substitution integrals is to try different substitutions until one leads to a simpler integrand. Another tip is to simplify the integrand before substituting, by using algebraic manipulation or trigonometric identities. It is also helpful to practice and become familiar with common substitutions and their corresponding integrals.

• Calculus and Beyond Homework Help
Replies
4
Views
739
• Calculus and Beyond Homework Help
Replies
3
Views
582
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
762
• Calculus and Beyond Homework Help
Replies
2
Views
703
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
783
• Calculus and Beyond Homework Help
Replies
2
Views
543
• Calculus and Beyond Homework Help
Replies
22
Views
1K