Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
How to simplify the question in the orange box and integrate to get the answer??
Date Posted: 3 years ago
Views: 220
can double check your question? I differentiate the ans I do not get the integral
Ohno let me try to check with my teachers :O thanks!
This indefinite integral does not exist.
This indefinite integral does not exist.
The answer is correct. The question has an error. The ln has been typed there by mistake.
It should be : ∫ 1/(eˣ + 1) dx
Use the method of integration by substitution.
Let u = eˣ
du/dx = eˣ → "dx = 1/eˣ du" → " dx = 1/u du"
Then,
∫ 1/(eˣ + 1) dx
= ∫ 1/(u + 1) · (1/u) du
= ∫ 1 / [u(u+1)] du
= ∫ (u + 1 - u) / [u(u+1)] du
= ∫ ( (u+1)/[u(u+1)] - u/[u(u+1)] ) du
= ∫ ( 1/u - 1/(u+1) ) du
《This is basically splitting it up into partial fractions》
= ln u - ln (u + 1)
= ln (eˣ) - ln (eˣ + 1)
《This is the answer key provided, but it is actually possible to simplify further》
= x ln e - ln (eˣ + 1)
= x - ln (eˣ + 1) + c
Where c is an arbitrary constant.
The answer is correct. The question has an error. The ln has been typed there by mistake.
It should be : ∫ 1/(eˣ + 1) dx
Use the method of integration by substitution.
Let u = eˣ
du/dx = eˣ → "dx = 1/eˣ du" → " dx = 1/u du"
Then,
∫ 1/(eˣ + 1) dx
= ∫ 1/(u + 1) · (1/u) du
= ∫ 1 / [u(u+1)] du
= ∫ (u + 1 - u) / [u(u+1)] du
= ∫ ( (u+1)/[u(u+1)] - u/[u(u+1)] ) du
= ∫ ( 1/u - 1/(u+1) ) du
《This is basically splitting it up into partial fractions》
= ln u - ln (u + 1)
= ln (eˣ) - ln (eˣ + 1)
《This is the answer key provided, but it is actually possible to simplify further》
= x ln e - ln (eˣ + 1)
= x - ln (eˣ + 1) + c
Where c is an arbitrary constant.
See 1 Answer
Answered in the question's comments section.
OHHh i see thank you so much~
360 Tutoring Program