How do you find the antiderivative of #int 1/(2x^2-x-3) dx#?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

2

This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Learn more about featured answers

Steve Share
Nov 15, 2016

Answer:

# int 1/(2x^2-x-3) dx = 1/7 ln| (x-2)/(x+3/2) | + C#

Explanation:

First we complete the square of the denominator:
Let # I = int 1/(2x^2-x-3) dx #

# :. I = int 1/(2{x^2-1/2x-3/2}) dx #
# :. I = 1/2 int 1/({(x-1/4)^2 - (1/4)^2-3}) dx#
# :. I = 1/2 int 1/({(x-1/4)^2 - 1/16 - 3}) dx#
# :. I = 1/2 int 1/({(x-1/4)^2 - 49/16}) dx#
# :. I = 1/2 int 1/({(x-1/4)^2 - (7/4)^2}) dx#

Let #u = x-1/4 => (du)/dx=1 #, Applying this substitution we get:
# :. I = 1/2 int 1/(u^2 - (7/4)^2) du#

Now a standard integral (that should be learnt if you are studying College Maths) is:
# int 1/(x^2-a^2)dx = 1/(2a)ln| (x-a)/(x+a) | #

And so,
# :. I = 1/2 1/(2(7/4))ln| (u-7/4)/(u+7/4) | #
# :. I = 1/7 ln| ((x-1/4)-7/4)/((x-1/4)+7/4) | #
# :. I = 1/7 ln| (x-2)/(x+3/2) | #

Was this helpful? Let the contributor know!
1500
Add an answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer
Impact of this question
Answer impact map
20 views around the world
You can reuse this answer
Creative Commons License