# Integration by substitution

As part of my Calculus 2 class, I have created a practice exam to help my students study.  In doing so, I felt that providing the questions and the solutions in a blog series would be helpful not just to my students, but also to anyone that may be taking, or teaching, calculus 2.  I hope you find the following solutions helpful in your studies.

You can also view me presenting the solution on YouTube at

## Find \begin{align*} \int_{0}^{1}x^{3}(x^{4}+2)^{3}dx. \end{align*}

First we find the anti-derivatives
\begin{align*} \int x^{3}(x^{4}+2)^{3}dx. \end{align*}

Here, we can simplify and integrate, or we can use substitution. I will use substitution with $$i=x^{4}+2$$, then $$di=4x^{3}dx$$ so $$dx=\frac{di}{4x^{3}}$$. We then substitute in $$i$$ and $$dx$$ and get
\begin{align*} \int x^{3}(x^{4}+2)^{3}dx&=\int x^{3}i^{3}\frac{di}{4x^{3}} \\ &=\frac{1}{4}\int i^{3}di \\ &=\frac{1}{4}\frac{i^{4}}{4}+c \\ &=\frac{(x^{4}+2)^{4}}{16}+c. \end{align*}

Using this anti-derivative, we can then find that
\begin{align*} \int_{0}^{1}x^{3}(x^{4}+2)^{3}dx&=\frac{(x^{4}+2)^{4}}{16}|_{0}^{1} \\ &=\frac{3^{4}}{16}-\frac{2^{4}}{16}=\frac{65}{16}. \end{align*}

## Other Resources

I have it linked above, but you can find the practice exam here.  You can find the other solutions to the practice exam in the other posts available here.

If you found the post, or the YouTube video helpful, please like the post or the video.  Also remember to follow the blog and subscribe to the YouTube channel.  By doing so you will not only be able to find future content, you will also make it easier for other students to find the content when they are studying for Calculus.  Thank you.

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