Product Rule and Quotient Rule

Introduction

Taking the derivatives of simple functions (i.e. polynomials) is easy using the power rule. For example, if

The method of  $u$-substitution is used to simplify the function you are integrating so that you can easily recognize it's antiderivative.

This method is closely related to the chain rule for derivatives.

One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master  $u$-substitution is to work out as many problems as possible. This will help you:

(1) understand the  $u$-substitution method and

(2) correctly identify the necessary substitution.

NOTE: After you plug-in your substitution, all of the  $x$'s in your integral should be gone. The only variables remaining in your integral should be  $u$'s.

Warm-Up

Evaluate the following indefinite integrals.

1)   $\int (8x+5)e^{4x^{2}+5x+3}~dx$

Solution:
Let  $u=4x^{2}+5x+3.$  Then,  $du=(8x+5)~dx.$
Plugging these into our integral, we get  $\int e^{u}~du,$  which we know how to integrate.
So, we get
${\begin{array}{rcl}\displaystyle {\int (8x+5)e^{4x^{2}+5x+3}~dx}&=&\displaystyle {\int e^{u}~du}\\&&\\&=&\displaystyle {e^{u}+C}\\&&\\&=&\displaystyle {e^{4x^{2}+5x+3}+C.}\\\end{array}}$

Final Answer:
$e^{4x^{2}+5x+3}+C$

2)   $\int {\frac {x}{\sqrt {1-2x^{2}}}}~dx$

Solution:
Let  $u=1-2x^{2}.$  Then,  $du=-4x~dx.$  Hence,  ${\frac {du}{-4}}=x~dx.$
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int {\frac {x}{\sqrt {1-2x^{2}}}}~dx}&=&\displaystyle {\int -{\frac {1}{4}}~u^{-{\frac {1}{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{2}}u^{\frac {1}{2}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C.}\\\end{array}}$

Final Answer:
$-{\frac {1}{2}}{\sqrt {1-2x^{2}}}+C$

3)   $\int {\frac {\sin(\ln x)}{x}}~dx$

Solution:
Let  $u=\ln(x).$  Then,  $du={\frac {1}{x}}~dx.$
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int {\frac {\sin(\ln x)}{x}}~dx}&=&\displaystyle {\int \sin(u)~du}\\&&\\&=&\displaystyle {-\cos(u)+C}\\&&\\&=&\displaystyle {-\cos(\ln x)+C.}\\\end{array}}$

Final Answer:
$-\cos(\ln x)+C$

4)   $\int xe^{x^{2}}~dx$

Solution:
Let  $u=x^{2}.$  Then,  $du=2x~dx$  and  ${\frac {du}{2}}=x~dx.$
Plugging these into our integral, we get
${\begin{array}{rcl}\displaystyle {\int xe^{x^{2}}~dx}&=&\displaystyle {\int {\frac {1}{2}}e^{u}~du}\\&&\\&=&\displaystyle {{\frac {1}{2}}e^{u}+C}\\&&\\&=&\displaystyle {{\frac {1}{2}}e^{x^{2}}+C.}\\\end{array}}$

Final Answer:
${\frac {1}{2}}e^{x^{2}}+C$

Exercise 1

Evaluate the indefinite integral  $\int {\frac {2}{y^{2}+4}}~dy.$

First, we factor out  $4$  out of the denominator.

So, we have

${\begin{array}{rcl}\displaystyle {\int {\frac {2}{y^{2}+4}}~dy}&=&\displaystyle {{\frac {1}{4}}\int {\frac {2}{{\frac {y^{2}}{4}}+1}}~dy}\\&&\\&=&\displaystyle {{\frac {1}{2}}\int {\frac {1}{({\frac {y}{2}})^{2}+1}}~dy.}\\\end{array}}$

Now, we use  $u$-substitution. Let  $u={\frac {y}{2}}.$

Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du={\frac {1}{2}}~dy}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 2~du=dy.}

Plugging these into our integral, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {2}{y^{2}+4}}~dy}&=&\displaystyle {{\frac {1}{2}}\int {\frac {2}{u^{2}+1}}~du}\\&&\\&=&\displaystyle {\int {\frac {1}{u^{2}+1}}~du}\\&&\\&=&\displaystyle {\arctan(u)+C}\\&&\\&=&\displaystyle {\arctan {\bigg (}{\frac {y}{2}}{\bigg )}+C.}\\\end{array}}}

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {2}{y^{2}+4}}~dy=\arctan {\bigg (}{\frac {y}{2}}{\bigg )}+C.}

Exercise 2

Evaluate the indefinite integral  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx.}

Let  $u=5+\sin(x).$  Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\cos(x)~dx.}

Plugging these into our integral, we get

${\begin{array}{rcl}\displaystyle {\int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx}&=&\displaystyle {\int {\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {-{\frac {1}{u}}+C}\\&&\\&=&\displaystyle {-{\frac {1}{5+\sin(x)}}+C.}\end{array}}$

So, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {\cos(x)}{(5+\sin x)^{2}}}~dx=-{\frac {1}{5+\sin(x)}}+C.}

Exercise 3

Evaluate the indefinite integral  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {x+5}{2x+3}}~dx.}

Here, the substitution is not obvious.

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=2x+3.}   Then,  $du=2~dx$  and  ${\frac {du}{2}}=dx.$

Now, we need a way of getting rid of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x+5}   in the numerator.

Solving for  $x$  in the first equation, we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x={\frac {1}{2}}u-{\frac {3}{2}}.}

Plugging these into our integral, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {x+5}{2x+3}}~dx}&=&\displaystyle {\int {\frac {({\frac {1}{2}}u-{\frac {3}{2}})+5}{2u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{2}}\int {\frac {{\frac {1}{2}}u+{\frac {7}{2}}}{u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{4}}\int {\frac {u+7}{u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{4}}\int 1+{\frac {7}{u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{4}}(u+7\ln |u|)+C}\\&&\\&=&\displaystyle {{\frac {1}{4}}(2x+3+7\ln |2x+3|)+C.}\\\end{array}}}

So, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int {\frac {x+5}{2x+3}}~dx={\frac {1}{4}}(2x+3+7\ln |2x+3|)+C.}

Exercise 4

Evaluate the indefinite integral  $\int {\frac {x^{2}+4}{x+2}}~dx.$

Let  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x+2.}   Then,  $du=dx.$

Now, we need a way of replacing  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x^{2}+4.}

If we solve for  $x$  in our first equation, we get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=u-2.}

Now, we square both sides of this last equation to get  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x^{2}=(u-2)^{2}.}

Plugging in to our integral, we get

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int {\frac {x^{2}+4}{x+2}}~dx}&=&\displaystyle {\int {\frac {(u-2)^{2}+4}{u}}~du}\\&&\\&=&\displaystyle {\int {\frac {u^{2}-4u+4+4}{u}}~du}\\&&\\&=&\displaystyle {\int {\frac {u^{2}-4u+8}{u}}~du}\\&&\\&=&\displaystyle {\int u-4+{\frac {8}{u}}~du}\\&&\\&=&\displaystyle {{\frac {u^{2}}{2}}-4u+8\ln |u|+C}\\&&\\&=&\displaystyle {{\frac {(x+2)^{2}}{2}}-4(x+2)+8\ln |x+2|+C.}\\\end{array}}}

So, we have

$\int {\frac {x^{2}+4}{x+2}}~dx={\frac {(x+2)^{2}}{2}}-4(x+2)+8\ln |x+2|+C.$