# Integration by Completing the Square

By changing the square, we may rewrite any quadratic polynomial ax² + bx + c in the form

$a\left( {{u^2} + {k}} \right),\;\;\text{where}\;\;u = x + \frac{b}{{2a}},\;\;k = \frac{{4ac - {b^2}}}{{4{a^2}}}.$

Completing the square helps when quadratic functions are involved in the integrand.

When the integrand is a rational function with a quadratic expression in the denominator, we can use the following table integrals:

$\int {\frac{{dx}}{{\sqrt {{x^2} \pm {a^2}} }}} = \ln \left| {x + \sqrt {{x^2} \pm {a^2}} } \right|,$
$\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \arcsin \frac{x}{a},$
$\int {\frac{{dx}}{{{a^2} + {x^2}}}} = \frac{1}{a}\arctan \frac{x}{a},$
$\int {\frac{{dx}}{{{a^2} - {x^2}}}} = \frac{1}{{2a}}\ln \left| {\frac{{a + x}}{{a - x}}} \right|.$

Certain other types of integrals involving quadratic functions can be evaluated using trigonometric and hyperbolic substitutions.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the integral $\int {\frac{{dx}}{{{x^2} - 5x + 7}}}.$

### Example 2

Evaluate the integral $\int {\frac{{dx}}{{{x^2} - x + 2}}}.$

### Example 3

Evaluate the integral $\int {\frac{{dx}}{{{x^2} + 10x + 26}}}.$

### Example 4

Evaluate the integral $\int {\frac{{dx}}{{5 - 4x - {x^2}}}}.$

### Example 5

Find the integral $\int {\frac{{dx}}{{\sqrt {{x^2} + x - 2} }}}.$

### Example 6

Find the integral $\int {\frac{{dx}}{{8 - 2x - {x^2}}}}.$

### Example 1.

Find the integral $\int {\frac{{dx}}{{{x^2} - 5x + 7}}}.$

Solution.

We complete the square in the denominator:

${x^2} - 5x + 7 = {x^2} - 2 \cdot \frac{5}{2}x + {\left( {\frac{5}{2}} \right)^2} - {\left( {\frac{5}{2}} \right)^2} + 7 = {\left( {x - \frac{5}{2}} \right)^2} + 7 - \frac{{25}}{4} = {\left( {x - \frac{5}{2}} \right)^2} + \frac{{28 - 25}}{4} = {\left( {x - \frac{5}{2}} \right)^2} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}.$

Make the substitution $$u = x - \frac{5}{2}.$$ Then $$du = dx,$$ and the integral in $$u-$$terms becomes

$\int {\frac{{dx}}{{{x^2} - 5x + 7}}} = \int {\frac{{dx}}{{{{\left( {x - \frac{5}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = \int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} .$

We can easily evaluate the last integral:

$\int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} = \frac{1}{{\frac{{\sqrt 3 }}{2}}}\arctan \frac{u}{{\frac{{\sqrt 3 }}{2}}} + C = \frac{2}{{\sqrt 3 }}\arctan \frac{{x - \frac{5}{2}}}{{\frac{{\sqrt 3 }}{2}}} + C = \frac{2}{{\sqrt 3 }}\arctan \frac{{2x - 5}}{{\sqrt 3 }} + C.$

### Example 2.

Evaluate the integral $\int {\frac{{dx}}{{{x^2} - x + 2}}}.$

Solution.

We complete the square in the quadratic expression:

${x^2} - x + 2 = {x^2} - 2 \cdot x \cdot \frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} - {\left( {\frac{1}{2}} \right)^2} + 2 = {\left( {x - \frac{1}{2}} \right)^2} - \frac{1}{4} + 2 = {\left( {x - \frac{1}{2}} \right)^2} + \frac{7}{4} = {\left( {x - \frac{1}{2}} \right)^2} + {\left( {\frac{{\sqrt 7 }}{2}} \right)^2}.$

Making the substitution

$u = x - \frac{1}{2},\;\;du = dx,$

we find the integral:

$\int {\frac{{dx}}{{{x^2} - x + 2}}} = \int {\frac{{dx}}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 7 }}{2}} \right)}^2}}}} = \int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 7 }}{2}} \right)}^2}}}} = \frac{1}{{\frac{{\sqrt 7 }}{2}}}\arctan \frac{u}{{\frac{{\sqrt 7 }}{2}}} + C = \frac{2}{{\sqrt 7 }}\arctan \frac{{x - \frac{1}{2}}}{{\frac{{\sqrt 7 }}{2}}} + C = \frac{2}{{\sqrt 7 }}\arctan \frac{{2x - 1}}{{\sqrt 7 }} + C.$

### Example 3.

Evaluate the integral $\int {\frac{{dx}}{{{x^2} + 10x + 26}}}.$

Solution.

We complete the square in the denominator:

${x^2} + 10x + 26 = {x^2} + 10x + 25 + 1 = {\left( {x + 5} \right)^2} + 1,$

so the integral is written as

$I = \int {\frac{{dx}}{{{x^2} + 10x + 26}}} = \int {\frac{{dx}}{{{{\left( {x + 5} \right)}^2} + 1}}} .$

Using the substitution

$u = x + 5,\;\;du = dx,$

we have

$I = \int {\frac{{du}}{{{u^2} + 1}}} = \arctan u + C = \arctan \left( {x + 5} \right) + C.$

### Example 4.

Evaluate the integral $\int {\frac{{dx}}{{5 - 4x - {x^2}}}}.$

Solution.

First we complete the square in the denominator:

$5 - 4x - {x^2} = 5 - \left( {{x^2} + 4x} \right) = 9 - \left( {{x^2} + 4x + 4} \right) = 9 - {\left( {x + 2} \right)^2} = {3^2} - {\left( {x + 2} \right)^2}.$

Then the integral becomes

$I = \int {\frac{{dx}}{{5 - 4x - {x^2}}}} = \int {\frac{{dx}}{{{3^2} - {{\left( {x + 2} \right)}^2}}}} .$

Using

$u = x + 2,\;\;du = dx,$

we obtain

$I = \int {\frac{{du}}{{{3^2} - {u^2}}}} = \frac{1}{{2 \cdot 3}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right| + C = \frac{1}{6}\ln \left| {\frac{{3 + x + 2}}{{3 - \left( {x + 2} \right)}}} \right| + C = \frac{1}{6}\ln \left| {\frac{{5 + x}}{{1 - x}}} \right| + C.$

### Example 5.

Find the integral $\int {\frac{{dx}}{{\sqrt {{x^2} + x - 2} }}}.$

Solution.

Complete the square in the denominator:

${x^2} + x - 2 = \left( {{x^2} + x + \frac{1}{4}} \right) - \frac{1}{4} - 2 = {\left( {x + \frac{1}{2}} \right)^2} - \frac{9}{4} = {\left( {x + \frac{1}{2}} \right)^2} - {\left( {\frac{3}{2}} \right)^2}.$

Making the substitution

$u = x + \frac{1}{2},\;\;du = dx,$

we obtain

$\int {\frac{{dx}}{{\sqrt {{x^2} + x - 2} }}} = \int {\frac{{dx}}{{\sqrt {{{\left( {x + \frac{1}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2}} }}} = \int {\frac{{du}}{{\sqrt {{u^2} - {{\left( {\frac{3}{2}} \right)}^2}} }}} = \ln \left| {u + \sqrt {{u^2} - {{\left( {\frac{3}{2}} \right)}^2}} } \right| + C = \ln \left| {x + \frac{1}{2} + \sqrt {{x^2} + x - 2} } \right| + C.$

### Example 6.

Find the integral $\int {\frac{{dx}}{{8 - 2x - {x^2}}}}.$

Solution.

Completing the square in the denominator yields:

$8 - 2x - {x^2} = 8 - \left( {{x^2} + 2x} \right) = 9 - \left( {{x^2} + 2x + 1} \right) = {3^2} - {\left( {x + 1} \right)^2}.$

Hence, the integral can written in the form

$I = \int {\frac{{dx}}{{8 - 2x - {x^2}}}} = \int {\frac{{dx}}{{{3^2} - {{\left( {x + 1} \right)}^2}}}} .$

Making the substitution

$u = x + 1,\;\;du = dx,$

we have

$I = \int {\frac{{dx}}{{{3^2} - {{\left( {x + 1} \right)}^2}}}} = \int {\frac{{du}}{{{3^2} - {u^2}}}} = \frac{1}{{2 \cdot 3}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right| + C = \frac{1}{6}\ln \left| {\frac{{3 + x + 1}}{{3 - \left( {x + 1} \right)}}} \right| + C = \frac{1}{6}\ln \left| {\frac{{4 + x}}{{2 - x}}} \right| + C.$

See more problems on Page 2.