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Integration of Particular Functions




Integration of Particular Functions


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

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

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

D. $\int \frac {1}{\sqrt {a^2 - x^2}} dx =  \sin ^{-1} (\frac {x}{a}) + C$
Proof
Let $x = a sin\theta$. Then $dx = a cos \theta d\theta$
So,
$\int \frac {1}{\sqrt {a^2 - x^2}} dx $
$=\int \frac {a cos \theta}{\sqrt {a^2 - a^2 sin^2 \theta }}d\theta$
$=\int d\theta = \theta + C = \sin ^{-1} (\frac {x}{a}) + C$

E. $\int \frac {1}{\sqrt {a^2  + x^2}} dx =  ln |x + \sqrt {a^2  + x^2}|   + C$
Proof
Put $x =a tan \theta$ then $dx= a sec^2 \theta d\theta$
Therefore
$\int \frac {1}{\sqrt {a^2  + x^2}} dx$
$=\int sec \theta d\theta$
$=ln |sec \theta + tan \theta| + C$
Now
$tan \theta = \frac {x}{a}$
$sec \theta = \sqrt {1 + tan^2 \theta } = \sqrt {1 + \frac {x^2}{a^2}} $
Therefore
$=ln | \frac {x}{a} + \sqrt {1 + \frac {x^2}{a^2}}| + C_1 = ln |x + \sqrt {a^2  + x^2}|   - ln |a| +C_1 = ln |x + \sqrt {a^2  + x^2}|  + C$

F. $\int \frac {1}{\sqrt {x^2  - a^2}} dx =  ln |x + \sqrt {x^2  - a^2}|   + C$
Proof
Put $x =a sec \theta$ then $dx= a sec \theta tan \theta d\theta $
Therefore
$\int \frac {1}{\sqrt {x^2 -a^2}} dx $
$=\int sec \theta d\theta $
$=ln |sec \theta + tan \theta| + C $
Now
$sec \theta = \frac {x}{a} $
$tan \theta = \sqrt {sec^2 \theta -1 } = \sqrt {\frac {x^2}{a^2} -1} $
Therefore
$=ln | \frac {x}{a} + \sqrt {\frac {x^2}{a^2} -1}| + C_1 = ln |x + \sqrt {x^2  - a^2}|   - ln |a| +C_1 = ln |x + \sqrt {a^2  + x^2}|  + C$
The above formulas are very useful to solve the various types of Integration Problems

Problem Type I

$\int \frac {1}{ax^2 + bx + c} dx$
This integral can be converted to the form A,B, C given above and can be evaluated using the formula

Example 1
$\int \frac {1}{x^2 -6x + 25} dx$
Solution
$\int \frac {1}{x^2 -6x + 25} dx= \int \frac {1}{(x -3)^2 + 4^2} dx$
Substituting y=x-3, or dy=dx, we get
$=\int \frac {1}{y^2 + 4^2} dy$
Now this can be easily evaluated using the formula A with a =4
$\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$
Hence
$=\frac {1}{4} \tan ^{-1} (\frac {y}{4}) + C$
Substituting back the value of y
$=\frac {1}{4} \tan ^{-1} (\frac {x-3}{4}) + C$

Problem Type II

$\int \frac {1}{\sqrt {ax^2 + bx + c}} dx$
This integral can be converted to the form D,E,F given above and can be evaluated using the formula

Problem Type III

$\int \frac {px + q}{ax^2 + bx + c} dx$
Here we can decompose this integral using the below formula
$px+q = A \frac {d}{dx} (ax^2 + bx + c) + B$
Then integral is converted as
$=A \int \frac {\frac {d}{dx} (ax^2 + bx + c)} {ax^2 + bx + c} dx + B \int \frac {1}{ax^2 + bx + c} dx $

The first integral is of the form
$\int \frac {f'(x)}{f(x)} dx$ which can be easily evaluated
The second integral is of the form of problem type I
$\int \frac {1}{ax^2 + bx + c} dx $ which can be easily evaluated

Problem Type IV

$\int \frac {px + q}{\sqrt{ax^2 + bx + c}} dx$
Here we can decompose this integral using the below formula
$px+q = A \frac {d}{dx} (ax^2 + bx + c) + B$
Then integral is converted as
$=A \int \frac {\frac {d}{dx} (ax^2 + bx + c)} {ax^2 + bx + c} dx + B \int \frac {1}{\sqrt {ax^2 + bx + c}} dx $

The first integral is of the form
$\int \frac {f'(x)}{f(x)^{1/2}} dx$ which can be easily evaluated
The second integral is of the form of problem type II
$\int \frac {1}{\sqrt {ax^2 + bx + c}} dx$ which can be easily evaluated


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