Question 1 :
Integrate the following functions with respect to x :
ex log a ex
Solution :
∫[ex log a ex] dx
= ∫elog a^x ex dx
= ∫x ex dx
u = x, dv = ex
du = dx, v = ex
∫ udv = uv - ∫v du
∫x ex dx = x(ex) - ∫ ex dx
= xex - ex + c
= x (ex -1) + c
Question 2 :
Integrate the following functions with respect to x :
(3x + 4)√(3x + 7)
Solution :
∫(3x + 4)√(3x + 7) dx
Let t = 3x + 7
3x = t - 7
3x + 4 = t - 7 + 4 ==> t - 3
Differentiating with respect to "x", we get
dt = 3 dx ==> dx = dt/3
∫(3x + 4)√(3x + 7) dx = ∫(t - 3)√t (dt/3)
= (1/3)[∫ t√t dt - 3∫√t dt]
= (1/3)[∫ t3/2 dt - 3∫t1/2 dt]
= (1/3)[(2/5) t5/2 - 3 (2/3)t3/2]
= (1/3)[(2/5) (3x + 7)5/2 - 2 (3x + 7)3/2] + c
Question 3 :
Integrate the following functions with respect to x :
(81 + x + 41- x) / 2x
Solution :
∫(81 + x + 41- x) / 2x dx
= ∫(23(1+x) + 22(1-x) / 2x dx
= ∫(23+3x + 22-2x) / 2x dx
= ∫(23+3x-x + 22-2x-x) dx
= ∫(23+2x + 22-3x) dx
= (23+2x/2log 2) - (22-3x/3 log 2) + c
= (22+2x/log 2) - (22-3x/3 log 2) + c
Question 4 :
Integrate the following functions with respect to x :
1/(√(x + 3) - √(x + 4))
Solution :
∫ 1/(√(x + 3) - √(x + 4)) dx
= ∫(1/(√(x + 3) - √(x + 4)) dx
Rationalizing the denominator, we get
= ∫[(√(x + 3) + √(x + 4))/(x + 3) - (x + 4))]dx
= -[∫(√(x + 3) + √(x + 4)) dx
= - [∫(√(x + 3) dx + ∫√(x + 4)) dx]
= (2/3)(x + 3)3/2 + (2/3)(x + 4)3/2 + c
Question 5 :
Integrate the following functions with respect to x :
(x + 1)/(x + 2)(x + 3)
Solution :
∫ [(x + 1)/(x + 2)(x + 3)] dx
We have to use the concept partial fraction in order to integrate.
(x + 1)/(x + 2)(x + 3) = A/(x + 2) + B/(x + 3)
x + 1 = A(x + 3) + B(x + 2)
Plug x = -2
-1 = A ===> A = -1
Plug x = -3
-2 = -B ===> B = 2
∫(x+1)/(x+2)(x+3) dx = -∫1/(x + 2) dx + ∫ 2/(x + 3) dx
= - log (x + 2) + 2 log (x + 3) + c
= 2 log (x + 3) - log (x + 2) + c
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Aug 31, 24 09:05 PM
Aug 31, 24 08:51 PM
Aug 31, 24 06:59 PM