Algebraic Integrals Examples

Algebraic integrals depict a numerical value equal to the area under the graph of a function for a new function or to express an interval. They are often cumbersome to solve as they involve lengthy calculations. There are several ways of solving algebraic integrals where standard methods can be applied to reach the solutions quickly. 

These are some easy ways of obtaining integrals of various rational and irrational functions: 

1. Integrals of the form:

 ∫1/(ax2 + bx + c) dx or reducible to the form ∫1/(ax2 + bx + c) dx

We seek to express the term ax2 + bx + c as the sum or the difference of two squares for such integrals. Follow these steps:

• The coefficient of x2 is made a unity by dividing or multiplying it throughout 
• Then, add or subtract the square of half of the coefficient of x to express it in the form 

[(x + (b/2a))2 + (4ac-b2)/4a2].

• The obtained equation can be easily integrated after this, using the formula:

 ∫1/(x2 – a2) or ∫1/(a2 + x2) or ∫1/(a2 – x2)

2. Integrals of the form:

 ∫1/√(ax2 + bx + c) dx or reducible to the form ∫1/√(ax2 + bx + c) dx

For integrating such functions, follow the following steps:

• The coefficient of x2 is made a unity by dividing or multiplying it throughout
• Then, we add or subtract the square of half of the coefficient of x to express it in the form, that is, (1/2 coefficient of x)2 inside the square root to express the term within the square root in the form 

[(x + (b/2a))2 + (4ac-b2)/4a2] or (4ac-b2)/4a2] – (x + (b/2a))2.

• The obtained equation can be easily integrated after this, using the formula:

∫ 1/√(x2 – a2) or ∫ 1/√(a2 + x2) or ∫ 1/√(a2 – x2).

3. Integrals of the form:

∫ (px + q)/ √(ax2 + bx + c) dx

In this situation, too, we proceed along the same lines as in the previous one:

• The numerator is equal to the sum of constant times denominator differentiation plus a constant., that is,

write px + q = A(2ax + b) + C, where 2ax + b = d/dx (ax2 + bx + c) and A and C are temporary constants.

• The values of constants are computed by equating the coefficients of like powers of x on both sides
• The numerator (px + q) is replaced by A(2ax + b) + C in the given integral which would produce

∫ (px + q)/ √(ax2 + bx + c) dx = A ∫ (2ax + b)/√(ax2 + bx + c) dx + C ∫ 1/√(ax2 + bx + c) dx

• This can be integrated and then the values of the constants can be substituted.

4. Integrals of the form:

∫ (px + q)/(ax2 + bx + c) dx

• In such cases, the numerator is expressed as the sum of constant times differentiation of denominator and a constant, that is

px + q = A(2ax + b) + C, where 2ax + b = d/dx (ax2 + bx + c) and A and C are temporary constants.

• The values of the constants can be easily obtained by equating the coefficients with the same powers on both sides 
• Replace the numerator (px + q) by A(2ax + b) + C in the given integral which would produce

∫ (px + q)/(ax2 + bx + c) dx = A ∫ (2ax + b)/(ax2 + bx + c) dx + C ∫ 1/(ax2 + bx + c) dx

• The R.H.S. thus obtained can easily be integrated, and then the values of the constants as obtained above can be substituted.

5. Integrals of the form:

∫ √(ax2 + bx + c) dx

In such types of integrals, 

• The coefficient of x2 is made a unity by dividing it throughout to get x2 + b/a x + c/a
• The term is made a perfect square by subtracting and adding the square of half of the coefficient of x and obtaining (x + b/2a)2 + (4ac-b2)/4a2
• The integral reduces to one of the following forms:

∫√a2+x2 dx, ∫√a2-x2 dx, or ∫√x2-a2 dx.

• These integrals can be calculated by applying a direct formulae.

6. Integrals of the form:

∫ (px + q) √(ax2 + bx + c) dx

• First, we express (px + q) as px + q = A (2ax + b) + B, where A and B are constants
• Now, by equating the coefficients of x and constant terms on both sides, we can obtain the values of A and B
• By replacing (px + q) in the given integral by A (2ax + b) + B we obtain the following form:

∫ (px + q) √(ax2 + bx + c) dx = A ∫(2ax + b)√(ax2 + bx + c) dx + B ∫√(ax2 + bx + c) dx 

• These integrals thus obtained can be easily computed by applying direct formulae.

7. Integrals of the form:

∫ (px + q)/√(ax2 + bx + c) dx

• In such cases, the numerator is expressed as the sum of constant times differentiation of denominator and a constant, that is,

px + q = C1 (d/dx (ax2 + bx + c) + C2, where C1 and C2 are temporary constants.

• The values of the constants can be easily obtained by equating the coefficients of like powers on both sides
• Replace the numerator (px + q) by C1 (2ax + b) + C2 in the given integral which would produce 

∫ (px + q)/√(ax2 + bx + c) dx = C1 ∫ (2ax + b)/√(ax2 + bx + c) dx + C2 ∫ 1/√(ax2 + bx + c) dx

• The obtained equation on the right hand side can now easily be integrated, and then the constant values obtained can be substituted

Conclusion

Algebraic integrals are one of the most important and challenging concepts applied, especially in physics and mathematics. However, they are tedious and challenging to solve, requiring constant practice. These solved algebraic integrals examples will help you devise quicker strategies for solving them. The best way to improvise is to solve more and more questions.