i need to solve this integral:
dx/(9x^2+6x+2)
so i need to complete the square so i can do trig sub. and i dont get how to simplify the denominator.
in the book it says it should simplify so the denominator looks like this:
(3x+1)^2 + 1
i've never completed the square with a coefficient infront of the x^2 before so i need help please. thanks.
dx/(9x^2+6x+2)
so i need to complete the square so i can do trig sub. and i dont get how to simplify the denominator.
in the book it says it should simplify so the denominator looks like this:
(3x+1)^2 + 1
i've never completed the square with a coefficient infront of the x^2 before so i need help please. thanks.
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9x² + 6x + 2
= 9(x² + 2x/3 + 2/9)
= 9[(x + 1/3)² - 1/9 + 2/9]
= 9[(x + 1/3)² + 1/9]
= 9(x + 1/3)² + 1
= [3(x + 1/3)]² + 1
= (3x + 1)² + 1
You should then make the substitution tanu = (3x + 1)
(1/3)sec²u du = dx
and
1 + (3x + 1)² = 1 + tan² = sec²u. It is a straightforward integral after this.
= 9(x² + 2x/3 + 2/9)
= 9[(x + 1/3)² - 1/9 + 2/9]
= 9[(x + 1/3)² + 1/9]
= 9(x + 1/3)² + 1
= [3(x + 1/3)]² + 1
= (3x + 1)² + 1
You should then make the substitution tanu = (3x + 1)
(1/3)sec²u du = dx
and
1 + (3x + 1)² = 1 + tan² = sec²u. It is a straightforward integral after this.
-
go to wolfram alpha,
then type in "integrate" then write your function next,
it will compute the answer for you, then show you the steps you need to take to get there
here are examples
http://www.wolframalpha.com/examples/Calculus.html
then type in "integrate" then write your function next,
it will compute the answer for you, then show you the steps you need to take to get there
here are examples
http://www.wolframalpha.com/examples/Calculus.html