integration of 1/a^2+ x^2, 1/a^2- x^2, 1/x^2 -a^2

Integration of 1/a^2+ x^2, 1/a^2- x^2, 1/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are

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

$\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$

$\int \frac {1}{a^2 – x^2} dx = \frac {1}{2a} ln |\frac {a+x}{a-x}| + C$

Proof of integration of 1/a^2+ x^2

This can be derived using trigonometric substitution and trigonometric identities

$\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$
Proof
Put $x =a \tan \theta$
then
$dx= a \sec^2 \theta d\theta$
Therefore
$\int \frac {1}{x^2 + a^2} dx$
$=\int \frac {a \sec^2 \theta}{a^2 \tan^2 \theta + a^2} d\theta$
Now $\sec^2 \theta = \tan^2 \theta + 1$

Therefore
$=\frac {1}{a} \int d\theta= \frac {1}{a} \theta + C $

Substituting back the values

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

Proof of integration of 1/a^2 – x^2

This can be derived using integration by partial fractions

$\int \frac {1}{a^2 – x^2} dx = \frac {1}{2a} ln |\frac {a+x}{a-x}| + C$
Proof
Using partial fraction technique
$\frac {1}{a^2 – x^2} = \frac {A}{a-x} + \frac {B}{a+x}$

Solving for A and B , we get

$\frac {1}{a^2 – x^2} =\frac {1}{2a}[ \frac {1}{a-x} + \frac {1}{a+x}]$
So
$\int \frac {1}{a^2 – x^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{a-x} dx + \int \frac {1}{x+a}]$
$= \frac {1}{2a}[-ln |a-x| + ln |a+x| + C$
$=\frac {1}{2a} ln |\frac {a+x}{a-x}| + C$

Proof of integration of 1/x^2 – a^2

This can be derived using integration by partial fractions

$\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$
Proof

Using partial fraction technique
$\frac {1}{x^2 – a^2} = \frac {A}{x-a} + \frac {B}{x+a}$

Solving for A and B , we get

$\frac {1}{x^2 – a^2} =\frac {1}{2a}[ \frac {1}{x-a} – \frac {1}{x+a}]$
So
$\int \frac {1}{x^2 – a^2} dx $
$=\frac {1}{2a}[ \int \frac {1}{x-a} dx – \int \frac {1}{x+a}]$
$= \frac {1}{2a}[ln |x-a| – ln |x+a| + C$
$=\frac {1}{2a} ln |\frac {x-a}{x+a}| + C$

Integration Based on the above

$\int \frac {1}{ax^2 + bx + c} dx$

This integral can be converted to the form A,B, C given above using completing the square method and can be evaluated using the formula

Solved Examples

Question 1:

$$ \int \frac{1}{x^2 + 4} \, dx $$

Solution

In this case, $ a^2 = 4 $, so $ a = 2 $. Using the formula:

$$ \int \frac {1}{x^2 + a^2} \, dx = \frac {1}{a} \tan ^{-1} \left( \frac {x}{a} \right) + C $$

we get:

$$ \int \frac{1}{x^2 + 4} \, dx = \frac {1}{2} \tan ^{-1} \left( \frac {x}{2} \right) + C $$

Question 2:

$$ \int \frac{1}{9x^2 + 1} \, dx $$

Solution

Here, we need to bring the denominator to the form $ x^2 + a^2 $. We can rewrite $ 9x^2 + 1 $ as $ (3x)^2 + 1^2 $. Thus, $ a = 1 $ and the integral becomes:

$$ \int \frac{1}{9x^2 + 1} \, dx = \int \frac{1}{(3x)^2 + 1^2} \, dx $$

Using the formula:

$$ \int \frac {1}{x^2 + a^2} \, dx = \frac {1}{a} \tan ^{-1} \left( \frac {x}{a} \right) + C $$

we get:

$$ \int \frac{1}{9x^2 + 1} \, dx = \frac {1}{3} \tan ^{-1} (3x) + C $$

Question 3:

$$ \int \frac{1}{9 – x^2} \, dx $$

Solution

In this case, $ a^2 = 9 $, so $ a = 3 $. Using the formula, the integral becomes:

$$ \int \frac{1}{9 – x^2} \, dx = \frac{1}{2 \times 3} \ln \left| \frac{3 + x}{3 – x} \right| + C $$
$$ = \frac{1}{6} \ln \left| \frac{3 + x}{3 – x} \right| + C $$

Question 4:

$$ \int \frac{1}{1 – 4x^2} \, dx $$
Solution

Here, we can rewrite $ 1 – 4x^2 $ as $ (1)^2 – (2x)^2 $. Thus, $ a = 1 $ and the integral becomes:

$$ \int \frac{1}{1 – 4x^2} \, dx = \int \frac{1}{1^2 – (2x)^2} \, dx $$

Using the formula, we get:

$$ \int \frac{1}{1 – 4x^2} \, dx = \frac{1}{2} \ln \left| \frac{1 + 2x}{1 – 2x} \right| + C $$

Question 5

$$\int \frac {1}{x^2 + 2x + 7} \; dx$$

Solution

To integrate the function $\int \frac {1}{x^2 + 2x + 7} \, dx$, we can use a method involving completing the square and then applying a standard integral formula. The goal is to transform the quadratic in the denominator into a form that resembles the sum of a square and a constant.

Complete the Square:
The complete square form of $x^2 + 2x + 7$ is $(x + 1)^2 + 6$.

Apply the Integral Formula:
The integral now takes the form $\int \frac {1}{(x + 1)^2 + 6} \, dx$. This resembles the integral of the form $\int \frac {1}{u^2 + a^2} \, du$, which is solved as $\frac {1}{a} \tan ^{-1} \left( \frac {u}{a} \right) + C$.

Let $u = x + 1$ and $a^2 = 6$ (so $a = \sqrt{6}$), then du =dx

$$ \int \frac {1}{x^2 + 2x + 7} \, dx = \int \frac {1}{u^2 + (\sqrt{6})^2} \, du $$
Applying the formula
$= \frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{u}{\sqrt{6}(}\right) + C$
Substituting back
$=\frac{1}{\sqrt{6}} \tan^{-1}\left(\frac{\sqrt{6}(x + 1)}{6}\right) + C $$