Class 12 Maths Integrals Integrals of Some Particular Functions

Integrals of Some Particular Functions

                       Class_12_Maths_Integrals_Formulas_Of_Some_Particular_Integrals      

To find the integral ʃ dx/(ax2 + bx + c)

We write ax2 + bx + c = a[x2 + bx/a + c/a] = a[(x + b/2a)2 + (c/a – b2/4a2)]

Put x + b/2a = t so that dx = dt and write c/a – b2/4a2 = ±k2

Now, ʃ dx/(ax2 + bx + c) = (1/a) * ʃ dt/(t2 ± k2)

If c/a – b2/4a2 is +ve then we put + k2 and if If c/a – b2/4a2 is +ve then we put - k2

 

Problem: Find the integral

1/(9x2 + 6x + 5)

Solution:

1/(9x2 + 6x + 5) = 1/{(3x + 1)2 + 22}

Let 3x + 1 = t

=> 3 dx = dt

=> dx = dt/3

So, ʃ dx/(9x2 + 6x + 5) = (1/3) * ʃ dt/(t2 + 22)

                                       = (1/3) * [(1/2) * tan-1(t/2)] + C

                                       = (1/6) * tan-1{(3x + 1)/2)} + C

To find the integral of type ʃ dx/√(ax2 + bx + c):

To integrate ʃ dx/√(ax2 + bx + c), we do the same process as to find the integral of type ʃ dx/(ax2 + bx + c) and then apply the standard formula to obtain the solution.

Problem:  Find the integral

1/√(x2 + 2x + 2)

Solution:

1/√(x2 + 2x + 2) = 1/√{(x + 1)2 + 12}

Let x + 1 = t

=> dx = dt

So, ʃ dx/√(x2 + 2x + 2) = ʃ dt/√(t2 + 12)

                                       = log[t + √(t2 + 1)] + C

                                       = log[(x + 1) + √{(x + 1)2 + 1)}] + C

                                       = log[(x + 1) + √(x2 + 2x + 2)] + C

 

To find the integral of type ʃ [(px + q)/(ax2 + bx + c)] dx where p, q, r and s are constant:

 To integrate ʃ [(px + q)/(ax2 + bx + c)] dx, we have to find real numbers A and B such that

      px + q = A * d(ax2 + bx + c)/dx + B

=> px + q = A(2ax + b) + B

To determine the value of A and B, we equate from both sides the coefficients of x and the constant terms. After finding A and B, the integral is reduced to one of the known forms.

 

Problem: Find the integral

 (4x + 1)/(2x2 + x - 3)     

Solution:

Let 4x + 1 = A * d(2x2 + x - 3)/dx + B

=> 4x + 1 = A(4x + 1) + B

=> 4x + 1 = 4Ax + A + B

Equating the coefficients of x and constant term on both sides, we get

4A = 4

=> A = 1

A + B = 1

=> 1 + B = 1

=> B = 0

Let 2x2 + x – 3 = t

=> (4x + 1)dx = dt

Now, ʃ [(4x + 1)/(2x2 + x - 3)]dx = ʃ dt/t

                                                       = log t + C

                                                       = log(2x2 + x - 3) + C

 

To find the integral of type ʃ [(px + q)/√(ax2 + bx + c)] dx where p, q, r and s are constant:

 To integrate ʃ [(px + q)/√(ax2 + bx + c)] dx, we do the same process as to find the integral of type

 ʃ [(px + q)/(ax2 + bx + c)] dx and transform the integral into known standard forms.

Problem: Find the integral

(5x + 3)/√(x2 + 4x + 10)

Solution:

Let (5x + 3) = A * d(x2 + 4x + 10)/dx + B

=> (5x + 3) = A(2x + 4) + B

Equating the coefficients of x and constant term on both sides, we get

2A = 5

=> A = 5/2

And 4A + B = 3

=> B = -7

So, (5x + 3) = 5(2x + 4)/2 – 7

Now, ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = ʃ [{5(2x + 4)/2 – 7}/√(x2 + 4x + 10)]dx

                                                             = (5/2) * ʃ [(2x + 4)/√(x2 + 4x + 10)]dx - 7 * ʃ dx/√(x2 + 4x + 10)

Let I1 = ʃ [(2x + 4)/√(x2 + 4x + 10)]dx and I2 = ʃ dx/√(x2 + 4x + 10)

So, ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5I1/2 – 7I2    ……………..1

Now, consider I1 = ʃ [(2x + 4)/√(x2 + 4x + 10)]dx

Let x2 + 4x + 10 = t

=> (2x + 4)dx = dt

So, I1 = ʃ dt/√t

=> I1 = 2√t

=> I1 = 2√(x2 + 4x + 10)

Again consider I2 = ʃ dx/√(x2 + 4x + 10)

=> I2 = ʃ dx/√(x2 + 4x + 4 + 6)

=> I2 = ʃ dx/√{(x + 2)2 + 6}

=> I2 = ʃ dx/√{(x + 2)2 + (√6)2}

=> I2 = log{(x + 2) + (x2 + 4x + 10)}

From equation 1, we get

      ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5[2√(x2 + 4x + 10)]/2 – 7 * log{(x + 2) + (x2 + 4x + 10)}

=> ʃ [(5x + 3)/√(x2 + 4x + 10)]dx = 5 * √(x2 + 4x + 10) – 7 * log{(x + 2) + (x2 + 4x + 10)}

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