Integration Worksheet4 solution8

In this page integration worksheet4 solution8 we are going to see solution of some practice question from the worksheet of integration.

Question 28

Integrate the following with respect to x,  1/[√(ax + b) - √(ax + c)]

Solution:

Now we are going to multiply by the conjugate of the denominator

       = ∫ 1/[√(ax + b) - √(ax + c)] dx

       = ∫  1/[√(ax + b) - √(ax + c)] x [√(ax + b) + √(ax + c)] dx

       = ∫ [√(ax + b) + √(ax + c)]/[√(ax + b)² - √(ax + c)²] dx

       = ∫ [√(ax + b) + √(ax + c)]/[ax + b - ax - c] dx

       = ∫ [√(ax + b) + √(ax + c)]/(b-c) dx

       = [1/(b-c)]∫ [(ax + b)^(1/2) + (ax + c)^(1/2)] dx

       = [1/(b-c)] [(ax + b)^(3/2)/(3a/2) + (ax + c)^(3/2)/(3a/2)] + C

       = [1/(b-c)] [(2/3a)(ax + b)^(3/2) + (2/3a)(ax + c)^(3/2)] + C

       = [2/3a(b-c)] [(ax + b)^(3/2) + (ax + c)^(3/2)] + C


Question 29

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

Solution:

      = ∫ (x + 1) √(x + 3) dx

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

t = x + 3

x = t - 3

differentiating with respect to x on both sides

dt = 1 dx

      = ∫(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(2/3)[t^(3/2)] + C

      = [(2/5)(x+3)^(5/2)] - (4/3)[(x+3)^(3/2)] + C


Question 30

Integrate the following with respect to x,(x - 4) √(x + 7)

Solution:

      = ∫(x - 4) √(x + 7) dx

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

t = x + 7

x = t - 7

differentiating with respect to x on both sides

dt = 1 dx

      = ∫(t - 7 - 4)√t  dt

      = ∫(t - 11)t^(1/2)  dt

      = ∫(t^[1+(1/2)] - 11 t^(1/2))  dt

      = ∫[(t^(3/2)]/(3/2) - 11 t^(1/2))  dt

      = [(t^(5/2)]/(5/2) - 11 [t^(3/2)]/(3/2) + C

      = [(2/5)(t^(5/2)] - 11(2/3)[t^(3/2)] + C

      = [(2/5)(x+7)^(5/2)] - (22/3)[(x+7)^(3/2)] + C

integration worksheet4 solution8 integration worksheet4 solution8