Continuing with Integrals of the Form:& Partial Fractions

Chapter 7.3 & 7.4

a2 −u2 ,  u2 −a2 ,  a2 +u2

We use inverse substitution with x = g(t), and dx = g’(t)dtto transform:

a2 −u2

u2 −a2

  a2 +u2

u =asinθ 1−sin2θ =cos2θ

u =asecθ sec2 θ −1=tan2θ

 u =atanθ tan2 θ +1=sec2θ

f (x)dx = f g(t)( )g'(t)dt∫∫

Evaluate:

Let

We can take advantage of:

By looking at how the radical expression changes:

x =2secθ

sec2 θ −1=tan2θ

dx =2secθ tanθdθ

x2 −4 = 2secθ( )2 −4

= 4 sec2θ − 4

= 4 sec2θ − 1( )

= 4 tan2θ

=2 tanθ

x2 −4x

dx∫

Chapter 7.3 & 7.4

March 29, 2007

Making the substitution into:

Let

We get:

To integrate, we use one of our trig identities again.

x =2secθdx =2secθ tanθdθ x2 −4 =2 tanθ

2 tanθ2secθ

2secθ tanθdθ∫ = 2 tan2θdθ∫

2 tan2 θdθ∫ =2 sec2θ −1( )∫ dθ

x2 −4x

dx∫

=2 tanθ −θ( )

Back to x?:

Let

We use the right triangle and our original substitution.

Using the Pythagorean Theorem, we find the third side:

Note this was our original expression.

Using the triangle and the expression:

Our answer in terms of x:

? = x2 −4

2 tanθ −θ( )

2x2 −42

−cos−1 2x

⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜

⎞

⎠⎟

=2 tanθ −θ( )x2 −4x

dx∫

x =2secθdx =2secθ tanθdθ x2 −4 =2 tanθ

x =2secθ ⇒x

2= secθ ⇒

2

x= cosθ

= x2 − 4 − 2 cos−1 2

x⎛⎝⎜

⎞⎠⎟

x2 −4

Another form:

Let

We use the right triangle and our original substitution.

Or using the tangent expression:

Our answer becomes:

x2 −4x

dx∫

x =2secθdx =2secθ tanθdθ x2 −4 =2 tanθ

x =2secθ ⇒x

2= secθ ⇒

2

x= cosθ

= x2 − 4 − 2cos−1 2

x⎛⎝⎜

⎞⎠⎟

= x2 − 4 − 2 tan−1 x2 − 4

2

⎛

⎝⎜

⎞

⎠⎟

x2 −4x2 −4 =2 tanθ ⇒

x2 − 4

2= tanθ

=2 tanθ −θ( )

Suppose you used the substitution:

9

2θ +

sin2θ2

⎛⎝⎜

⎞⎠⎟

ln secθ + tanθ

cotθ −secθ

x =3sinθ

Complete the problem if the integration resulted in:

How would your solution change if the substitution was instead?x = 2 tanθ

Determine the integral obtained when using the appropriate trig. change of variable:

5 −4x−x2( )12∫  dx

1

5x2 + 2( )∫  dx

1

x2 x2 + 4( )∫  dx

7.4 Integrating Rational Functions or“Undoing” Addition!

First check is to be sure the Rational Function is in reduced form. In reduced form, the degree of the numerator is less than the degree of the denominator. If not, the first step is to simplify the expression using LONG division……..

So for example, the rational function in the integral below is NOT in reduced form.

We can simply the expression by dividing:

In reduced form we have:

x2 + 5x−2x+ 3

⎛

⎝⎜⎞

⎠⎟dx∫

x + 3 x2 + 5x−2x

x2 + 3x−x2 − 3x2x−2

+  2

2x + 6−2x − 6−8

x + 2 −8

x+ 3⎛⎝⎜

⎞⎠⎟dx∫ =

x2

2+ 2x − 8 ln x + 3

Example:

The rational function is NOT in reduced form. So divide:

In reduced form we have:

x4 + 4 5x5 +        5x

5x

5x5  +         20x

5x−15xx4 + 4

⎛⎝⎜

⎞⎠⎟dx∫

5x5 + 5xx4 + 4

⎛

⎝⎜⎞

⎠⎟dx∫

−5x5  −         20x

−15x

=5x2

2−

15x

x4 + 4⎛⎝⎜

⎞⎠⎟dx∫

Example:

To Complete the integration, use u substitution and the arctangent formula!

5x−15xx4 + 4

⎛⎝⎜

⎞⎠⎟dx∫

5x5 + 5xx4 + 4

⎛

⎝⎜⎞

⎠⎟dx∫

=5x2

2−

15x

x4 + 4⎛⎝⎜

⎞⎠⎟dx∫

=5x2

2−

15

2

2x

x2( )2

+ 22

⎛

⎝⎜⎜

⎞

⎠⎟⎟dx∫ =

5x2

2−

15

2

1

u( )2

+ 22

⎛

⎝⎜

⎞

⎠⎟du∫

u =x2

du=2xdx

=5x2

2−

15

2

1

2⎛⎝⎜

⎞⎠⎟

tan−1 u

2⎛⎝⎜

⎞⎠⎟

=5x2

2−

15

4tan−1 x2

2

⎛

⎝⎜⎞

⎠⎟

The Arctangent formula (also see day 9 notes)

The “new” formula:

When to use? If the polynomial in the denominator does not have real roots (b2-4ac < 0) then the integral is an arctangent, we complete the square and integrate….

For example:

1

u2 + a2 du=1atan−1 u

a⎛⎝⎜

⎞⎠⎟∫

1

2x2 +10x+13dx∫ =

11

22x + 5( )

2+ 1( )

dx∫ =21

2x + 5( )2

+ 1dx∫

u =2x+ 5du=2dx

=tan−1 2x + 5( )

What if the polynomial has real roots?

That means we can factor the polynomial and “undo” the addition! To add fractions we find a common denominator and add: we’ll work the other way….

The denominator of our rational function factors into (t - 4)(t +1) So in our original “addition,” the fractions were of the form:

1

t 2 + 3t−4dt∫

a

b+cd

=ad+ cbbd

ad + cbbd

=ab

+cd

1

t 2 + 3t−4=

At−4

+Bt+1

What if the polynomial has real roots?

We now need to solve for A and B. Multiplying both sides by the denominator:

Choose t to be -1 and substituting into the equation above, we get 1 = -5B or B = -1/5

And with t = 4: 1 = 5A, or A = 1/5

1

t 2 + 3t−4dt∫

1

t 2 + 3t−4=

At−4

+Bt+1

1 =A(t+1) + B(t−4)

Integrating with the Partial Fraction Decomposition:

Our Rational function can be written as:

And the integral becomes:

Which integrates to:

1

t 2 + 3t−4dt∫

1

t 2 + 3t−4=

15

t−4+

−15

t+1

1

t 2 + 3t−4dt∫ =

15

1t−4

dt∫ + −15

1t+1

dt∫

1

5ln t −4 −

15ln t+1 =

1

5lnt − 4

t +1

Examples:

x−9x2 + 3x−10

dx∫

x3 + x2 + 6x2 +1∫  dx

−2x − 6

x2 − 2x − 3∫  dx