There are some important integration formulas that are applied for integrating many other standard integrals. In this article, we will look at the integrals of these particular functions.

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## Integral of Some Particular Functions

Look at the following integration formulas

- âˆ« dx / (x
^{2}â€“ a^{2}) = 1/2a log |(x â€“ a) / (x + a)| + C - âˆ« dx / (a
^{2}â€“ x^{2}) = 1/2a log |(a + x) / (a â€“ x)| + C - âˆ« dx / (x
^{2}+ a^{2}) = 1/a tan^{â€“1}(x/a) + C - âˆ« dx / âˆš (x
^{2}â€“ a^{2}) = log |x + âˆš (x^{2}â€“ a^{2})| + C - âˆ« dx / âˆš (a
^{2}â€“ x^{2}) = sin^{â€“1}(x/a) + C - âˆ« dx / âˆš (x
^{2}+ a^{2}) = log |x + âˆš (x^{2}+ a^{2})| + C

## Proof of the Above Six Standard Integration Formulas

### 1. âˆ« dx / (x^{2} â€“ a^{2}) = 1/2a log |(x â€“ a) / (x + a)| + C

We know that,

1 / (x^{2} â€“ a^{2}) = 1 / (x â€“ a) (x + a) = 1/2a [(x + a) â€“ (x â€“ a) / (x â€“ a) (x + a)]

= 1/2a [1/(x â€“ a) â€“ 1/(x + a)]

Therefore,

âˆ« dx / (x^{2} â€“ a^{2}) = 1/2a [âˆ« dx / (x â€“ a) – âˆ« dx / (x + a)]

= 1/2a [log |(x â€“ a) â€“ log |(x + a)] + C

= 1/2a log |(x â€“ a) / (x + a)| + C

**Browse more Topics Under Integrals**

- Fundamental Theorem of Calculus
- Introduction to Integration
- Properties of Indefinite Integrals
- Properties of Definite Integrals
- Definite Integral as a Limit of a Sum
- Integration by Partial Fractions
- Integration by Parts
- Integration by Substitutions
- Integral of Some Particular Functions
- Integral of the Type e^x[f(x) + f'(x)]dx

### 2. âˆ« dx / (a^{2} â€“ x^{2}) = 1/2a log |(a + x) / (a â€“ x)| + C

We know that,

1 / (a^{2} â€“ x^{2}) = 1 / (a â€“ x) (a + x) = 1/2a [(a + x) + (a â€“ x) / (a â€“ x) (a + x)]

= 1/2a [1/(a â€“ x) + 1/(a + x)]

Therefore,

âˆ« dx / (a^{2} â€“ x^{2}) = 1/2a [âˆ« dx / (a â€“ x) + âˆ« dx / (a + x)]

= 1/2a [â€“ log |(a â€“ x) + log |(a + x)] + C

= 1/2a log |(a + x) / (a â€“ x)| + C

### 3. âˆ« dx / (x^{2} + a^{2}) = 1/a tan^{â€“1} (x/a) + C

Letâ€™s substitute x = a tan t, so we have dx = a sec^{2} t dt. Therefore,

âˆ« dx / (x^{2} + a^{2}) = âˆ« [(a sec^{2} t dt) / (a^{2} tan^{2} t + a^{2})]

Solving this, we get,

âˆ« dx / (x^{2} + a^{2}) = 1/a âˆ« dt = t/a + C

Re-substituting the value of t, we get

âˆ« dx / (x^{2} + a^{2}) = 1/a tan^{â€“1} (x/a) + C

#### 4. âˆ« dx / âˆš (x^{2} â€“ a^{2}) = log |x + âˆš (x^{2} â€“ a^{2})| + C

Letâ€™s substitute x = a sec t, so that dx = a sec t tan t dt. Therefore,

âˆ« dx / âˆš (x^{2} â€“ a^{2}) = âˆ« a sec t tan t dt / âˆš (a^{2} sec^{2} t â€“ a^{2})

Solving this, we get,

âˆ« dx / âˆš (x^{2} â€“ a^{2}) = âˆ« sec t dt = log |sec t + tan t| + C_{1
}Re-substituting the value of t, we get

âˆ« dx / âˆš (x^{2} â€“ a^{2}) = log |(x/a) + âˆš [(x^{2} â€“ a^{2}) / a^{2}]| + C_{1
}= log |x + âˆš(x^{2} â€“ a^{2})| â€“ log |a| + C_{1
}= log |x + âˆš(x^{2} â€“ a^{2})| + C â€¦ where C = C_{1} â€“ log |a|

### 5. âˆ« dx / âˆš (a^{2} â€“ x^{2}) = sin^{â€“1} (x/a) + C

Letâ€™s substitute x = a sin t, so that dx = a cos t dt. Therefore,

âˆ« dx / âˆš (a^{2} â€“ x^{2}) = âˆ« a cos t dt / âˆš (a^{2} â€“ a^{2 }sin^{2} t)

Solving this, we get,

âˆ« dx / âˆš (a^{2} â€“ x^{2}) = âˆ« t dt = t + C

Re-substituting the value of t, we get

âˆ« dx / âˆš (a^{2} â€“ x^{2}) = sin^{â€“1} (x/a) + C

### 6. âˆ« dx / âˆš (x^{2} + a^{2}) = log |x + âˆš (x^{2} + a^{2})| + C

Letâ€™s substitute x = a tan t, so that dx = a sec^{2} t dt. Therefore,

âˆ« dx / âˆš (x^{2} + a^{2}) = âˆ« a sec^{2} t dt / âˆš (a^{2} tan^{2} t + a^{2})

Solving this, we get,

âˆ« dx / âˆš (x^{2} â€“ a^{2}) = âˆ« sec t dt = log |sec t + tan t| + C_{1
}Re-substituting the value of t, we get

âˆ« dx / âˆš (x^{2} â€“ a^{2}) = log |(x/a) + âˆš [(x^{2} + a^{2}) / a^{2}]| + C_{1
}= log |x + âˆš(x^{2} + a^{2})| â€“ log |a| + C_{1
}= log |x + âˆš(x^{2} + a^{2})| + C â€¦ where C = C_{1} â€“ log |a|

Now, letâ€™s apply these standard integration formulas to obtain formulae that are applied directly to evaluate other integrals.

**7. Integral âˆ« dx / (ax**^{2} + bx + c)

^{2}+ bx + c)

We can write,

ax^{2} + bx + c = a [x^{2} + (b/a)x + (c/a)]

= a [(x + b/2a)^{2} + (c/a â€“ b^{2}/4a^{2})]

Now, letâ€™s substitute (x + b/2a) = t, so that dx = dt. Also, substitute (c/a â€“ b^{2}/4a^{2}) = Â±k^{2}. Therefore,

ax^{2} + bx + c = a (t^{2} Â± k^{2}) â€¦ where the + or â€“ depends on the sign of (c/a â€“ b^{2}/4a^{2}).

Hence,

âˆ« dx / (ax^{2} + bx + c) = 1/a âˆ« dt / (t^{2} Â± k^{2})

This can be evaluated using one / more of the six integration formulas shown above. Remember, you can also solve âˆ« dx / âˆš (ax^{2} + bx + c) in a similar manner.

### 8. Integral âˆ« [(px + q) / (ax^{2} + bx + c)] dx, where p, q, a, b, and c are constants.

To solve this, we must find constants A and B such that,

(px + q) = A d/dx (ax^{2} + bx + c) + B = A (2ax + b) + B

To determine â€˜Aâ€™ and â€˜Bâ€™, we equate from both sides the coefficients of x and the constant terms. â€˜Aâ€™ and â€˜Bâ€™ are thus obtained and hence the integral is reduced to one of the known forms. Letâ€™s understand with the help of some examples:

## Solved Problems for You

### Example 1: Find âˆ« [(x + 2) / (2x^{2} + 6x + 5)] dx

To solve this equation, we express

(x + 2) = A d/dx (2x^{2} + 6x + 5) + B = A (4x + 6) + B

Therefore, x + 2 = 4Ax + 6A + B

Next, letâ€™s equate the coefficients of â€˜xâ€™ and the constant terms. We have,

4A = 1 and 6A + B = 2

On solving them, we get

A = Â¼ and B = Â½

Hence, we have

âˆ« [(x + 2) / (2x^{2} + 6x + 5)] dx = Â¼ âˆ« [(4x + 6) / (2x^{2} + 6x + 5)] dx + Â½ âˆ« dx / (2x^{2} + 6x + 5)

= Â¼ âˆ« I_{1} + Â½ âˆ« I_{2
}Now, letâ€™s solve I_{1} and I_{2} separately.

#### Solving I_{1}

Letâ€™s substitute (2x^{2} + 6x + 5) = t, so that (4x + 6) dx = dt. Therefore,

I_{1} = âˆ« [(4x + 6) / (2x^{2} + 6x + 5)] dx = âˆ« dt/t = log |t| + C_{1
}Or, I_{1} = log |(2x^{2} + 6x + 5)| + C_{1} â€¦ (1.1)

#### Solving I_{2}

I_{2} = âˆ« dx / (2x^{2} + 6x + 5) = Â½ âˆ« dx / (x^{2} + 3x + 5/2) = Â½ âˆ« dx / [(x + 3/2)^{2} + (1/2)^{2}]

Now, letâ€™s substitute (x + 3/2) = t, so that dx = dt. Therefore,

I_{2} = Â½ âˆ« dt / [t^{2} + (1/2)^{2}]

Using the six integration formulas shown above, we get

I_{2} = (1/(2 x Â½) tan^{â€“1} 2t + C_{2} = tan^{â€“1} 2 (x + 3/2) + C_{2} = tan^{â€“1} (2x + 3) + C_{2} â€¦ (1.2)

Using (1.1) and (1.2), we get

âˆ« [(x + 2) / (2x^{2} + 6x + 5)] dx = Â¼ log |2x^{2} + 6x + 5| + Â½ tan^{â€“1} (2x + 3) + C

Where, C = C_{1}/4 + C_{2}/2

### Example 2: Find the integral of (x + 3) /Â âˆš (5 – 4x + x^{2}) with respect to x.

Solution: We can express,

x + 3 = A d/dx (5 â€“ 4x + x^{2}) + B = A (â€“ 4 â€“ 2x) + B

Equating the coefficients, we get

A = â€“ Â½ and B = 1

Therefore, âˆ« [(x + 3) / âˆš (5 â€“ 4x + x^{2})] dx = â€“ Â½ âˆ« [(â€“ 4 â€“ 2x) / âˆš (5 â€“ 4x + x^{2})] dx + âˆ« dx / âˆš (5 â€“ 4x + x^{2})

= â€“ Â½ I_{1} + I_{2} â€¦ (a)

#### Solving I_{1}

Letâ€™s substitute (5 â€“ 4x + x^{2}) = t, so that (â€“ 4 â€“ 2x) dx = dt. Therefore,

I_{1} = âˆ« [(â€“ 4 â€“ 2x) / âˆš (5 â€“ 4x + x^{2})] dx = âˆ« dt / âˆš t = 2 âˆš t + C_{1
}= 2 âˆš (5 â€“ 4x + x^{2}) + C_{1} â€¦ (b)

#### Solving I_{2}

I_{2} = âˆ« dx / âˆš (5 â€“ 4x + x^{2}) = âˆ« dx / âˆš [9 â€“ (x + 2)^{2}]

Now, letâ€™s substitute (x + 2) = t, so that dx = dt. Therefore,

I_{2} = âˆ« dt / âˆš (3^{2} â€“ t^{2}) = sin^{â€“1} (t/3) + C_{2
}= sin^{â€“1} [(x + 2) / 3] + C_{2} â€¦ (c)

Substituting (b) and (c) in (a), we get

âˆ« [(x + 3) / âˆš (5 â€“ 4x + x^{2})] dx = â€“ Â½ I_{1} + I_{2
}= â€“ âˆš (5 â€“ 4x + x^{2}) + sin^{â€“1} [(x + 2) / 3] + C â€¦ where C = C_{2} = C_{1}/2.

**Question.** How can one derive integration formulas?

**Answer.** The fundamental use of integration is as a version of summing that is continuous. One can derive integral by viewing integration as essentially an inverse operation to differentiation. One can call it the Fundamental Theorem of Calculus. Integration formulas involve almost the inverse operation of differentiation.

**Question.** What exactly do we understand by integration?

**Answer.** Simply speaking, integration refers to the act of bringing together smaller components into a single system. This single system is such that it functions as one.

**Question.** Explain the application of integration in real life?

**Answer.** Integration has various uses in real life. It is important in the fields of engineering, physics, medical science, research analysis, and graphic designing.

**Question.** Explain the rules for integration?

**Answer.** The rules for integration are power rule, constant coefficient rule, sum rule, and difference rule. The power rule gives the indefinite integral of a variable raised to a power. The constant coefficient rule informs us about the indefinite integral of c. f(x). The sum rule tells us about integrating functions that are the sum of several terms. The difference rule deals with the difference between two or more terms.

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