## Integration Worksheet5 solution9

In this page integration worksheet5 solution9 we are going to see solution of some practice question from the worksheet of integration.

Question 30

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

Solution:

= ∫ (x + 1)√(2 x + 3) dx

let t = 2 x + 3

differentiating with respect to "x"

dt = (2 (1) + 0) dx

d t = 2 dx

(t - 3)/2 = x

= ∫ (x + 1)√(2 x + 3) dx

= ∫ [((t - 3)/2) + 1] √t (dt/2)

= ∫ (t - 1)/2) √t (dt/2)

= (1/4) ∫ (t^3/2  - t^1/2) dt

= (1/4) [t^5/2/(5/2)  - t^3/2/(3/2)] + C

= (1/4) [(2/5) t^5/2 - (2/3) t^3/2] + C

= (1/4) [(2/5) (2 x + 3)^5/2 - (2/3) (2 x + 3)^3/2] + C

Question 31

Integrate the following with respect to x, (3 x + 5)√(2 x + 1)

Solution:

= ∫ (3 x + 5)√(2 x + 1) dx

let t = 2 x + 1

differentiating with respect to "x"

dt = (2 (1) + 0) dx

d t = 2 dx

dx = dt/2

(t - 1)/2 = x

= ∫ (3 x + 5)√(2 x + 1) dx

= ∫ [(3(t - 1)/2) + 5] √t (dt/2)

= ∫ (3t - 3 + 10)/2) √t (dt/2)

= ∫ [(3t + 7)/2] √t (dt/2)

= (1/4) ∫ (3t^3/2  + 7t^1/2) dt

= (1/4) [3t^5/2/(5/2) + 7t^3/2/(3/2)] + C

= (1/4) [(6/5) t^5/2 + (14/3) t^3/2] + C

= (1/4) [(6/5) (2 x+ 1)^5/2 + (14/3) (2 x + 1)^3/2] + C

Question 32

Integrate the following with respect to x, (x² + 1)√(x + 1)

Solution:

= ∫ (x² + 1)√(x + 1) dx

let t = x + 1

differentiating with respect to "x"

dt = dx

(t - 1) = x

= ∫ (x² + 1)√(x + 1) dx

= ∫ [(t - 1)² + 1] √t dt

= ∫ [(t² + 1 + 2 t) + 1] √t dt

= ∫ (t² + 2 + 2 t) √t dt

= ∫ (t²√t + 2√t + 2 t√t) dt

= ∫ (t^5/2 + 2t^1/2 + 2 t^3/2) dt

=  [t^7/2/(7/2) + 2t^3/2/(3/2) + 2 t^5/2/(5/2)] + C

=  [(2/7) t^7/2 + (4/3)t^3/2 + (4/5) t^5/2] + C

=  [(2/7)(x + 1)^7/2+(4/3)(x + 1)^3/2+(4/5)(x + 1)^5/2] + C integration worksheet5 solution9 integration worksheet5 solution9