Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

The u-substitution

Basically, there are two techniques of integration: substitution and parts. We will learn the basic principles of substitution in this course; other calculus courses provide an in-depth discussion of this technique and that of parts.

You can check that $u$-identification is not the correct approach to integrating

${\displaystyle \int x\sqrt{x+1}dx\text{.}}$

Instead, we use substitution, that is we make a substitution ${\displaystyle u=f\left(x\right)}$ such that each term of the integrand function can be substituted to obtain an integral in terms of $u$ only. In this case, if ${\displaystyle u=x+1\text{,}}$ then ${\displaystyle du=dx\text{.}}$ To replace ${\displaystyle x\text{,}}$ we must see that ${\displaystyle u-1=x\text{.}}$ Hence, the integral is

${\displaystyle \int x\sqrt{x+1}dx=\int \left(u-1\right)\sqrt{u}du\text{.}}$

The next step is to integrate the integral obtained in terms of ${\displaystyle u\text{,}}$ assuming that this can be done. In this case, we find that

${\displaystyle \begin{array}{ll}{\displaystyle \int u^{1/2}\left(u-1\right)du}\hfill & ={\displaystyle \int u^{3/2}}-u^{1/2}du\hfill \\ \hfill & ={\displaystyle \frac{2u^{5/2}}{5}}-{\displaystyle \frac{2u^{3/2}}{3}}+C\text{.}\hfill \end{array}}$

Since the answer must be in terms of ${\displaystyle x\text{,}}$ we substitute once more, and we obtain

${\displaystyle \int x\sqrt{x+1}dx={\displaystyle \frac{2{\left(x+1\right)}^{5∕2}}{5}}-{\displaystyle \frac{2{\left(x+1\right)}^{3∕2}}{2}}+C\text{.}}$

The challenge with this method is to choose the function $u$ correctly. There is no “recipe” for doing this; it can only be done by practice—that is, experience. We may have to try different substitutions to get the correct one. In the discussion that follows, if more than one substitution is possible, we tend to present the most efficient one, but there is no one substitution that is better than another: as long as they are correct, any one is acceptable.

We can summarize the steps in the substitution method as follows.

Step 1

Make a substitution of ${\displaystyle u=f\left(x\right)}$ for $f(x)$ a function in the integrand function, or any other function that relates to the integrand function.

Step 2

Replace each term of the integrand function in terms of ${\displaystyle u\text{,}}$ including the differential.

Step 3

Integrate the resulting integral in terms of ${\displaystyle u\text{.}}$

Step 4

Substitute $f(x)$ for ${\displaystyle u\text{,}}$ and give the antiderivative in terms of the original variable ${\displaystyle x\text{.}}$

Example 6.21. In the integral

${\displaystyle \int x^5\sqrt[3]{x^3+2}dx,}$

we choose ${\displaystyle u=x^3+2\text{.}}$ Then ${\displaystyle du=3x^{2}dx\text{.}}$

We can replace ${\displaystyle x^2\sqrt[3]{x^3+2}dx}$ with ${\displaystyle \sqrt[3]{u}{\displaystyle \frac{1}{3}}du\text{,}}$ and ${\displaystyle x^3}$ by ${\displaystyle u-2=x^3\text{.}}$

With these substitutions, we get

${\displaystyle \begin{array}{ll}{\displaystyle \frac{1}{3}}{\displaystyle \int \left(u-2\right)u^{1/3}du}\hfill & ={\displaystyle \frac{1}{3}}{\displaystyle \int u^{4/3}-2u^{1/3}du}\hfill \\ \hfill & ={\displaystyle \frac{u^{7/3}}{7}}-{\displaystyle \frac{2u^{4/3}}{4}}+C\text{.}\hfill \end{array}}$

The antiderivative is

${\displaystyle {\displaystyle \frac{{\left(x^3+2\right)}^{7∕3}}{7}}-{\displaystyle \frac{{\left(x^3+2\right)}^{4∕3}}{2}}+C\text{.}}$

Example 6.22. We use substitution with the integral

${\displaystyle \int x^2\sqrt{x+1}dx\text{.}}$

Let ${\displaystyle u=x+1\text{.}}$ Then ${\displaystyle du=dx\text{,}}$ and ${\displaystyle {\left(u-1\right)}^2=x^2}$

${\displaystyle \begin{array}{ll}{\displaystyle \int {\left(u-1\right)}^2{u}^{1/2}du}\hfill & ={\displaystyle \int u^{5/2}-2u^{3/2}+u^{1/2}}\hfill \\ \hfill & ={\displaystyle \frac{2u^{7/2}}{7}}-{\displaystyle \frac{4u^{5/2}}{5}}+{\displaystyle \frac{2u^{3/2}}{3}}+C\text{.}\hfill \end{array}}$

Hence,

${\displaystyle \int x^2\sqrt{x+1}dx={\displaystyle \frac{2{\left(x+1\right)}^{7∕2}}{7}}-{\displaystyle \frac{4{\left(x+1\right)}^{5∕2}}{5}}+{\displaystyle \frac{2{\left(x+1\right)}^{3∕2}}{3}}+C\text{.}}$

Tutorial 1: Integration

Tutorial 2: Integration