## Integration Worksheet4 solution9

In this page integration worksheet4 solution9 we are going to see solution of some practice question from the worksheet of integration.

Question 31

Integrate the following with respect to x,(2 x + 1) √(2 x + 3)

Solution:

= ∫(2 x + 1) √(2 x + 3) dx

Let us consider the term which is inside the radical as "t"

t = 2x + 3

2x = t - 3

differentiating with respect to x on both sides

2dx = dt

dx = dt/2

= ∫(t - 3 + 1)√t  dt

= ∫(t - 2)t^(1/2)  dt

= ∫(t^[1+(1/2)] - 2 t^(1/2))  dt

= ∫[(t^(3/2)]/(3/2) - 2 t^(1/2))  dt

= [(t^(5/2)]/(5/2) - 2 [t^(3/2)]/(3/2) + C

= [(2/5)(t^(5/2)] - 2[t^(3/2)] + C

= [(2/5)(2x+3)^(5/2)] - 2[(2x+3)^(3/2)] + C

Question 32

Integrate the following with respect to x,(x + 1)/[(x + 2) (x + 3)]

Solution:

(x + 1)/[(x + 2) (x + 3)] = [A/(x + 2)] + [B/(x + 3)]

(x + 1)/[(x + 2) (x + 3)] = [A(x+3)+B(x+2)/(x + 2) (x + 3)]

Since the denominators of both sides fractions are same. We can cancel the denominators.

(x + 1)= A(x+3)+B(x+2)

put x = -3

-3 + 1 = A (-3+3) + B (-3 + 2)

-2 = B(-1)

B = 2

put x = -2

-2 + 1 = A (-2+3) + B (-2 + 2)

-1 = A(1)

A = -1

= ∫[-1/(x + 2) + 2/(x + 3)] dx

= ∫ [-1/(x + 2) dx + ∫ [2/(x + 3)] dx

= -1 ∫ [1/(x + 2) dx + 2 ∫ [1/(x + 3)] dx

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

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

Question 33

Integrate the following with respect to x,(x² + 1)/[(x - 2)(x + 2)(x² + 9)]

Solution:

(x² + 1)/[(x - 2)(x + 2)(x² + 9)]=[A/(x-2)]+[B/(x+2)]+[(Cx+D)/(x²+9)]

(x²+1)/[(x-2)(x+2)(x²+9)]

= A(x+2)(x²+9)+B(x-2)(x²+9)+[(Cx+D)(x+2)(x-2)/[(x+2)(x-2)(x²+9)]

Since the denominators are same for both the fractions we can cancel them

(x²+1)= A(x+2)(x²+9)+B(x-2)(x²+9)+[(Cx+D)(x+2)(x-2)]

put x = 2

4+1 = A(2+2)(4+9)+B(0)+(Cx+D)(0)

5 = 52 A

A = 5/52

put x = -2

4+1 = A(0)+B(-2-2)(4+9)+(Cx+D)(0)

4+1 = B(-4)(4+9)

5 = -52 B

B = -5/52

put x = 0

0+1=A(2)(9)+B(-2)(9)+[D(2)(-2)]

1=18A-18B-4D

1 = 18(5/52)-18(-5/52)-4D

1 = (90/52)+(90/52)-4D

1 = (180/52) - 4D

1 - (180/52) = - 4 D

(52-180)/52 = - 4 D

128/52 = 4 D

D = 128/52 (4)

D = 32/52

D = 8/13

put x = 1

2 = A(3)(10)+B(-1)(10)+[(C+D)(3)(-1)]

2 = 30A-10B-3[(C+D)]

2 = 30(5/52)-10(-5/52)-3C-3(8/13)

2=(150/52)+(50/52)-3C-(24/13)

2=(200/52) -(96/52) - 3 C

2=(200-96)/52 - 3 C

2 - (104/52) = -3C

(104-104)/52 = - 3C

-3 C = 0

C = 0

= ∫[A/(x-2)] dx+∫[B/(x+2)]dx +∫[(Cx+D)/(x²+9)]dx

= ∫(5/52)[1/(x-2)]dx-∫(5/52)[1/(x+2)]dx+∫(8/13)[1/(x²+9)]dx

= (5/52)∫[1/(x-2)]dx-(5/52)∫[1/(x+2)]dx+(8/13)∫[1/(x²+9)]dx

= (5/52) log (x - 2) - (5/52) log (x + 2) + (8/13)(1/3) tan⁻¹(x/3) + C

= (5/52) log (x - 2) - (5/52) log (x + 2) + (8/39) tan⁻¹(x/3) + C integration worksheet4 solution9 integration worksheet4 solution9