Long division & completing the square | Integration

What you’ll learn:

Other integration techniques to simplify integrals into recognizable forms
More integrals involving inverse trig functions

In this section, we introduce two more techniques to rearrange an integral into a more recognizable form, so that integration rules or $u$ -substitution can be used.

Long division
Completing the square

Long division

When integrating a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, try long division first.

$\int \frac{2 x ^{2} + 3 x + 5}{x + 1} d x$

Solution

With polynomial long division,

$\frac{2 x ^{2} + 3 x + 5}{x + 1} = 2 x + 1 + \frac{4}{x + 1}$

So the integral is

$\int (2 x + 1 + \frac{4}{x + 1}) d x$

which becomes

$\int (2 x + 1) d x + 4 \int \frac{1}{x + 1} d x$

Use the reverse power rule for both terms of the 1st integral, and use $u$ -substitution on the 2nd integral with

$u = x + 1$

$d u = d x$

$\frac{2 x ^{2}}{2} + x + 4 \int \frac{1}{u} d u$

$= x^{2} + x + 4 ln ∣ u ∣$

$= x^{2} + x + 4 ln ∣ x + 1∣ + C$

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

Solution

(spoiler)

With polynomial long division,

$\frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} = x - 1 + \frac{- x + 3}{x ^{2} + 1}$

So the integral becomes

$\int (x - 1 + \frac{- x + 3}{x ^{2} + 1}) d x$

The integral can be broken up into

$\int (x - 1) d x + \int \frac{- x + 3}{x ^{2} + 1} d x$

$= \frac{x ^{2}}{2} - x + \int \frac{- x + 3}{x ^{2} + 1} d x$

$u$ -substitution doesn’t quite work on the remaining integral yet but we can further separate the fraction into

$\frac{x ^{2}}{2} - x + \int \frac{- x}{x ^{2} + 1} d x + \int \frac{3}{x ^{2} + 1} d x$

For the 1st integral $\int \frac{- x}{x ^{2} + 1} d x$ , use $u$ -substitution with

$u = x^{2} + 1$

Then

$d u = 2 x d x$

$- \frac{1}{2} d u = - x d x$

Rewriting the integral in terms of $u$ ,

$\int \frac{- x}{x ^{2} + 1} d x = \int \frac{1}{u} \cdot (- \frac{1}{2}) d u$

$= - \frac{1}{2} \int \frac{1}{u} d u$

$= - \frac{1}{2} ln ∣ u ∣$

$= - \frac{1}{2} ln (x^{2} + 1)$

The 2nd integral $\int \frac{3}{x ^{2} + 1} d x$ appears to be similar to the derivative of $tan^{- 1} (x)$ . Then

$\int \frac{3}{x ^{2} + 1} d x$

$= 3 \int \frac{1}{x ^{2} + 1} d x$

$= 3 tan^{- 1} (x)$

Putting everything together,

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

$= \frac{x ^{2}}{2} - x - \frac{1}{2} ln (x^{2} + 1) + 3 tan^{- 1} (x) + C$

Completing the square

If the integrand has a quadratic expression in the denominator or under a square root and it’s not easily factorable, try completing the square.

$\int \frac{1}{x ^{2} + 4 x + 5} d x$

Solution

(spoiler)

Completing the square for $x^{2} + 4 x + 5$ ,

$\int \frac{1}{( x ^{2} + 4 x + 4 ) + 5 - 4} d x$

$= \int \frac{1}{( x + 2 ) ^{2} + 1} d x$

This resembles the integral

$\int \frac{1}{x ^{2} + 1} d x = tan^{- 1} (x) + C$

once we make the $u$ -substitution

$u = x + 2$

$d u = d x$

Rewriting the integral in terms of $u$ ,

$\int \frac{1}{( x + 2 ) ^{2} + 1} d x$

$= \int \frac{1}{u ^{2} + 1} d u$

$= tan^{- 1} (u)$

$= tan^{- 1} (x + 2) + C$

$\int \frac{1}{- 2 x - x ^{2}} d x$

Solution

(spoiler)

Completing the square for $- x^{2} - 2 x$ ,

$\int \frac{1}{- ( x ^{2} + 2 x + 1 ) + 1} d x$

$= \int \frac{1}{1 - ( x + 1 ) ^{2}} d x$

This resembles the integral

$\frac{1}{1 - x ^{2}} d x = sin^{- 1} (x) + C$

once we set

$u = x + 1$

$d u = d x$

Rewriting the integral in terms of $u$ ,

$\frac{1}{1 - u ^{2}} d u$

$= sin^{- 1} (u)$

$= sin^{- 1} (x + 1) + C$

$\int \frac{6}{x ^{2} + 4 x + 13} d x$

Solution

(spoiler)

Completing the square,

$\int \frac{6}{( x ^{2} + 4 x + 4 ) + 13 - 4} d x$

$\int \frac{6}{( x + 2 ) ^{2} + 9} d x$

This resembles the integral

$\int \frac{1}{x ^{2} + 1} d x = tan^{- 1} (x) + C$

But the $9$ in the denominator has to be a $1$ . To achieve this, multiply both the top and bottom by $1/9$ :

$\int \frac{6 \cdot \frac{1}{9}}{(( x + 2 ) ^{2} + 9 ) \cdot \frac{1}{9}} d x$

$\int \frac{2}{3} \cdot \frac{1}{\frac{( x + 2 ) ^{2}}{9} + 1} d x$

$\frac{( x + 2 ) ^{2}}{9}$ in the denominator can be turned into $(\frac{x + 2}{3})^{2}$ .

$\int \frac{2}{3} \cdot \frac{1}{( \frac{x + 2}{3} ) ^{2} + 1} d x$

Then make the $u$ -substitution

$u = \frac{x + 2}{3}$

$d u = \frac{1}{3} d x$

The integral rewritten in terms of $u$ is

$\int \frac{2}{u ^{2} + 1} d x$

$= 2 tan^{- 1} (u)$

$= 2 tan^{- 1} (\frac{x + 2}{3}) + C$

What you’ll learn:

Other integration techniques to simplify integrals into recognizable forms
More integrals involving inverse trig functions

In this section, we introduce two more techniques to rearrange an integral into a more recognizable form, so that integration rules or $u$ -substitution can be used.

Long division
Completing the square

Long division

When integrating a rational function where the degree of the numerator is greater than or equal to the degree of the denominator, try long division first.

$\int \frac{2 x ^{2} + 3 x + 5}{x + 1} d x$

Solution

With polynomial long division,

$\frac{2 x ^{2} + 3 x + 5}{x + 1} = 2 x + 1 + \frac{4}{x + 1}$

So the integral is

$\int (2 x + 1 + \frac{4}{x + 1}) d x$

which becomes

$\int (2 x + 1) d x + 4 \int \frac{1}{x + 1} d x$

Use the reverse power rule for both terms of the 1st integral, and use $u$ -substitution on the 2nd integral with

$u = x + 1$

$d u = d x$

$\frac{2 x ^{2}}{2} + x + 4 \int \frac{1}{u} d u$

$= x^{2} + x + 4 ln ∣ u ∣$

$= x^{2} + x + 4 ln ∣ x + 1∣ + C$

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

Solution

(spoiler)

With polynomial long division,

$\frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} = x - 1 + \frac{- x + 3}{x ^{2} + 1}$

So the integral becomes

$\int (x - 1 + \frac{- x + 3}{x ^{2} + 1}) d x$

The integral can be broken up into

$\int (x - 1) d x + \int \frac{- x + 3}{x ^{2} + 1} d x$

$= \frac{x ^{2}}{2} - x + \int \frac{- x + 3}{x ^{2} + 1} d x$

$u$ -substitution doesn’t quite work on the remaining integral yet but we can further separate the fraction into

$\frac{x ^{2}}{2} - x + \int \frac{- x}{x ^{2} + 1} d x + \int \frac{3}{x ^{2} + 1} d x$

For the 1st integral $\int \frac{- x}{x ^{2} + 1} d x$ , use $u$ -substitution with

$u = x^{2} + 1$

Then

$d u = 2 x d x$

$- \frac{1}{2} d u = - x d x$

Rewriting the integral in terms of $u$ ,

$\int \frac{- x}{x ^{2} + 1} d x = \int \frac{1}{u} \cdot (- \frac{1}{2}) d u$

$= - \frac{1}{2} \int \frac{1}{u} d u$

$= - \frac{1}{2} ln ∣ u ∣$

$= - \frac{1}{2} ln (x^{2} + 1)$

The 2nd integral $\int \frac{3}{x ^{2} + 1} d x$ appears to be similar to the derivative of $tan^{- 1} (x)$ . Then

$\int \frac{3}{x ^{2} + 1} d x$

$= 3 \int \frac{1}{x ^{2} + 1} d x$

$= 3 tan^{- 1} (x)$

Putting everything together,

$\int \frac{x ^{3} - x ^{2} + 2}{x ^{2} + 1} d x$

$= \frac{x ^{2}}{2} - x - \frac{1}{2} ln (x^{2} + 1) + 3 tan^{- 1} (x) + C$

Completing the square

If the integrand has a quadratic expression in the denominator or under a square root and it’s not easily factorable, try completing the square.

$\int \frac{1}{x ^{2} + 4 x + 5} d x$

Solution

(spoiler)

Completing the square for $x^{2} + 4 x + 5$ ,

$\int \frac{1}{( x ^{2} + 4 x + 4 ) + 5 - 4} d x$

$= \int \frac{1}{( x + 2 ) ^{2} + 1} d x$

This resembles the integral

$\int \frac{1}{x ^{2} + 1} d x = tan^{- 1} (x) + C$

once we make the $u$ -substitution

$u = x + 2$

$d u = d x$

Rewriting the integral in terms of $u$ ,

$\int \frac{1}{( x + 2 ) ^{2} + 1} d x$

$= \int \frac{1}{u ^{2} + 1} d u$

$= tan^{- 1} (u)$

$= tan^{- 1} (x + 2) + C$

$\int \frac{1}{- 2 x - x ^{2}} d x$

Solution

(spoiler)

Completing the square for $- x^{2} - 2 x$ ,

$\int \frac{1}{- ( x ^{2} + 2 x + 1 ) + 1} d x$

$= \int \frac{1}{1 - ( x + 1 ) ^{2}} d x$

This resembles the integral

$\frac{1}{1 - x ^{2}} d x = sin^{- 1} (x) + C$

once we set

$u = x + 1$

$d u = d x$

Rewriting the integral in terms of $u$ ,

$\frac{1}{1 - u ^{2}} d u$

$= sin^{- 1} (u)$

$= sin^{- 1} (x + 1) + C$

$\int \frac{6}{x ^{2} + 4 x + 13} d x$

Solution

(spoiler)

Completing the square,

$\int \frac{6}{( x ^{2} + 4 x + 4 ) + 13 - 4} d x$

$\int \frac{6}{( x + 2 ) ^{2} + 9} d x$

This resembles the integral

$\int \frac{1}{x ^{2} + 1} d x = tan^{- 1} (x) + C$

But the $9$ in the denominator has to be a $1$ . To achieve this, multiply both the top and bottom by $1/9$ :

$\int \frac{6 \cdot \frac{1}{9}}{(( x + 2 ) ^{2} + 9 ) \cdot \frac{1}{9}} d x$

$\int \frac{2}{3} \cdot \frac{1}{\frac{( x + 2 ) ^{2}}{9} + 1} d x$

$\frac{( x + 2 ) ^{2}}{9}$ in the denominator can be turned into $(\frac{x + 2}{3})^{2}$ .

$\int \frac{2}{3} \cdot \frac{1}{( \frac{x + 2}{3} ) ^{2} + 1} d x$

Then make the $u$ -substitution

$u = \frac{x + 2}{3}$

$d u = \frac{1}{3} d x$

The integral rewritten in terms of $u$ is

$\int \frac{2}{u ^{2} + 1} d x$

$= 2 tan^{- 1} (u)$

$= 2 tan^{- 1} (\frac{x + 2}{3}) + C$