Up until now, we have done the substitution of \(u\) and all the \(x\) terms have canceled. However, what do you do when they do not all cancel? The idea is to use the equation that you set up for \(u\) and get it in a form so that you can replace all factors involving \(x\) with factors involving only \(u\). Once you have done this, the only variable in the integral is \(u\) which allows you to evaluate the integral. Let's look at an example to see how to do this.
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Example
Evaluate \( \int{ x \sqrt{x+2} ~dx } \).
Final Answer
\(\displaystyle{ \frac{2}{5}(x+2)^{5/2} \frac{4}{3}(x+2)^{3/2}+C }\) 

Problem Statement
Evaluate \( \int{ x \sqrt{x+2} ~dx } \).
Solution
Let \(u=x+2 \to du=1~dx\).
\( \int{ x \sqrt{x+2} ~dx = \int{ x \sqrt{u} ~du } } \)
Notice that we are not able to cancel out the extra \(x\) in the integral and we can't do the integration until we have only \(u\)'s. So what we do is go back to the initial equation for \(u\), which is \(u=x+2\) and we solve for \(x\) giving us \(x=u2\). We use this to substitute back into the integral for \(x\) to get rid of it. This gives us
\( \int{ (u2)\sqrt{u} ~du } \), which we can now evaluate.
\( \int{ (u2)\sqrt{u} ~du } = \int{ u^{3/2}  2u^{1/2} ~du } = \)
\(\displaystyle{ \frac{u^{5/2}}{5/2}  2\frac{u^{3/2}}{3/2}+C }\)
Now we substitute back in for \(u\) and we are done.
Note: If we are asked to simplify or factor, our answer will be \(\displaystyle{ 2(x+2)^{3/2}\left[ (x+2)/5 2/3 \right] +C }\)
Final Answer
\(\displaystyle{ \frac{2}{5}(x+2)^{5/2} \frac{4}{3}(x+2)^{3/2}+C }\) 

Here are some practice problems involving this technique.
Practice
Unless otherwise instructed, evaluate these integrals using integration by substitution. Give your answers in simplified, factored form.
\(\displaystyle{ \int{ \frac{x}{\sqrt[4]{x+2}}~dx } }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ \frac{x}{\sqrt[4]{x+2}}~dx } }\) using integration by substitution. Give your answer in simplified, factored form.
Solution
video by PatrickJMT 

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\(\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }\) using integration by substitution. Give your answer in simplified, factored form.
Hint 

For his substitution, he used \(u=\sqrt[3]{x+4}\). That is a little more difficult than we would recommend. Try \(u=x+4\) and see if this is easier to evaluate.
Problem Statement
Evaluate \(\displaystyle{ \int{x\sqrt[3]{x+4}~dx} }\) using integration by substitution. Give your answer in simplified, factored form.
Hint
For his substitution, he used \(u=\sqrt[3]{x+4}\). That is a little more difficult than we would recommend. Try \(u=x+4\) and see if this is easier to evaluate.
Solution
video by PatrickJMT 

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\(\displaystyle{ \int{ x^3 \sqrt{1x^2} ~ dx } }\)
Problem Statement 

Evaluate \(\displaystyle{ \int{ x^3 \sqrt{1x^2} ~ dx } }\) using integration by substitution. Give your answer in simplified, factored form.
Final Answer 

\(\displaystyle{ \int{ x^3 \sqrt{1x^2} ~ dx } }\) \(\displaystyle{ = \frac{1}{15} (1x^2)^{3/2} (3x^2+2) + C }\)
Problem Statement
Evaluate \(\displaystyle{ \int{ x^3 \sqrt{1x^2} ~ dx } }\) using integration by substitution. Give your answer in simplified, factored form.
Solution
Notation Comment  Although his final answer is correct, he has some incorrect notation during the course of his solution. Notice that he doesn't include his constant of integration until the very end. To make the entire solution precisely correct, he needs to include the constant of integration in the step right after he does the actual integration. This is required since he writes equal signs between his steps. (This would also be required if he implied each step is equal to the previous one.) So don't do this or you may lose points for your work. However, as usual, check with your instructor to see what they require.
video by Integrals ForYou 

Final Answer
\(\displaystyle{ \int{ x^3 \sqrt{1x^2} ~ dx } }\) \(\displaystyle{ = \frac{1}{15} (1x^2)^{3/2} (3x^2+2) + C }\)
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Really UNDERSTAND Calculus
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Topics You Need To Understand For This Page 

[substitution from precalculus]  [differentials]  [basics of integrals]  [integration by substitution] 
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