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# Section 3.5: Additional Integration Techniques

## Integration By Parts

Integration by parts is an integration method which enables us to find antiderivatives of some new functions such as $$\ln(x)$$ as well as antiderivatives of products of functions such as $$x^2\ln(x)$$ and $$xe^x$$.

If the function we're trying to integrate can be written as a product of two functions, $$u$$, and $$dv$$, then integration by parts lets us trade out a complicated integral for hopefully simpler one.

#### Integration by Parts Formula

$\int u\, dv = uv-\int v\, du$

For definite integrals: $\int_a^b u\, dv = \left.uv\right]_a^b-\int_a^b v\, du$

#### Example 1

Integrate $$\int xe^x\, dx$$.

To use the Integration by Parts method, we break apart the product into two parts: $u=x \qquad\text{and}\qquad dv=e^x\, dx.$

We now calculate $$du$$, the derivative of $$u$$, and $$v$$, the integral of $$dv$$: $du=\left(\frac{d}{dx} x\right)\, dx \qquad\text{and}\qquad v= \int e^x\, dx = e^x.$

Using the Integration by Parts formula, $\int xe^x\, dx=uv-\int v\, du = xe^x - \int e^x\, dx.$

Notice the remaining integral is simpler that the original, and one which we can easily evaluate: $xe^x - \int e^x\, dx = xe^x-e^x+C.$

In the last example we could have chosen either $$x$$ or $$e^x$$ as our $$u$$, but had we chosen $$u=e^x$$, the second integral would have become messier, rather than simpler.

#### Rule of Thumb

When selecting the $$u$$ for Integration by Parts, select a logarithmic expression if one is present. If not, select an algebraic expression (like $$x$$ or $$dx$$).

(There is a larger decision tree that can be written down for choosing $$u$$ and $$dv$$, but since we're not looking at any trigonometric functions in this course the rule above is sufficient for the functions we're integrating.)

#### Example 2

Integrate $$\int\limits_1^4\, 6x^2\ln(x)\, dx$$.

Since this contains a logarithmic expression, we'll use it for our u: $u=\ln(x) \qquad \text{and} \qquad dv= 6x^2\, dx$

We now calculate $$du$$ and $$v$$: $du=\frac{1}{x}dx \qquad \text{and} \qquad v= \int 6x^2\, dx = 6\frac{x^3}{3}=2x^3$

Using the By Parts formula: $\int_1^4 6x^2\ln(x)\, dx = \left.2x^3\ln(x)\right]_1^4 - \int_1^4 6x^2\frac{1}{x}\, dx$

We can simplify the expression in the integral on the right: $\int_1^4 6x^2\ln(x)\, dx = \left.2x^3\ln(x)\right]_1^4 - \int_1^4 6x\, dx$

The remaining integral is a basic one we can now evaluate: $\int_1^4 6x^2\ln(x)\, dx = \left.2x^3\ln(x)\right]_1^4 - \left.3x^2\right]_1^4$

Finally, we can evaluate the expressions: \begin{align*} \int_1^4 6x^2\ln(x)\, dx=& \left(\left(2\cdot 4^3\ln(4)\right)-\left(2\cdot 1^3\ln(1)\right)\right)-\left(\left(3\cdot 4^2\right)-\left(3\cdot 1^2\right)\right)\\ =& 128\ln(4)-45\\ \approx & 132.446 \end{align*}

## Integration Using Tables of Integrals

There are many techniques of integration we will not be studying. Many of them lead to general formulas which can be compiled into a Table of Integrals – a type of cheat-sheet for integration.

For example, here are two entries you might find in a table of integrals:

#### Table of Integral Examples

\begin{align*} \int \frac{1}{x^2-a^2}\, dx =& \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C \\ \int \frac{1}{\sqrt{x^2+a^2}}\, dx =& \ln\left|x+\sqrt{x^2+a^2}\right|+C \end{align*}

#### Example 3

Integrate $$\int\frac{5}{x^2-9}\, dx$$.

This integral looks very similar to the form of the first integral in the examples table. By employing the rule that allows us to pull out constants, and by rewriting 9 as $$3^2$$, we can better see the match. $\int\frac{5}{x^2-9}\, dx = 5\int\frac{1}{x^2-3^2}\, dx$

Now we simply use the formula from the table, with $$a = 3$$. \begin{align*} \int\frac{5}{x^2-9}\, dx = & 5\int\frac{1}{x^2-3^2}\, dx \\ =& 5\left(\frac{1}{2\cdot 3}\ln\left|\frac{x-3}{x+3}\right|\right)+C \\ =& \frac{5}{6}\ln\left|\frac{x-3}{x+3}\right|+C \end{align*}

Sometimes we have to combine the table with other techniques we've learned, like substitution.

#### Example 4

Integrate $$\int\frac{x^2}{\sqrt{x^6+16}}\, dx$$.

This integral looks somewhat like the second integral in the example table, but the power of x is incorrect, and there is an x2 in the numerator which does not match. Trying to utilize this rule, we can try to rewrite the denominator to look like (something)$$^2$$. Luckily, $$x^6 = \left(x^3\right)^2$$. $\int\frac{x^2}{\sqrt{x^6+16}}\, dx = \int\frac{x^2}{\sqrt{\left(x^3\right)^2+16}}\, dx$

Now we can use substitution, letting $$u=x^3$$, so $$du=3x^2\, dx$$.

Making the subsitution: $\int\frac{x^2}{\sqrt{\left(x^3\right)^2+16}}\, dx = \int\frac{1}{\sqrt{u^2+16}}\, \frac{du}{3} = \frac{1}{3}\int\frac{1}{\sqrt{u^2+16}}\, du$

Now we can use the table entry: $\frac{1}{3}\int\frac{1}{\sqrt{u^2+16}}\, du = \frac{1}{3}\ln\left|u+\sqrt{u^2+16}\right|+C$

Undoing the substitution yields the final answer: $\int\frac{x^2}{\sqrt{x^6+16}}\, dx = \frac{1}{3}\ln\left|x^3+\sqrt{x^6+16}\right|+C$