# Thread: Integral with sub. method?

1. ## Integral with sub. method?

Can someone explain how to use the substitution method to find the integral of:

$\displaystyle \int_{-1}^{1} \frac{2x + 1}{x^2 + x + 1}dx$

2. Originally Posted by wizzler
Can someone explain how to use the substitution method to find the integral of:

$\displaystyle \int_{-1}^{1} \frac{2x + 1}{x^2 + x + 1}dx$
Let $\displaystyle t=x^2+x+1$ then $\displaystyle t'=2x+1$.

3. Originally Posted by ThePerfectHacker
Let $\displaystyle t=x^2+x+1$ then $\displaystyle t'=2x+1$.

can you show me all the steps, i.e. the whole solution? I seem to get like $\displaystyle \frac{t'}{t}$

4. Originally Posted by wizzler
can you show me all the steps, i.e. the whole solution? I seem to get like $\displaystyle \frac{t'}{t}$
That's what you should get!

$\displaystyle \int \frac{t^{\prime}}{t}dt = ln|t| + C$

-Dan

5. Originally Posted by topsquark
That's what you should get!

$\displaystyle \int \frac{t^{\prime}}{t}dt = ln|t| + C$

-Dan

oh right.. so then i can just put in 3 and 1 for the function because if x = 1 then t = 3; if x = -1 then t = 1 so I get: (Note it's a definite integral!!!)

$\displaystyle ln(3) - ln(1)$ and that is the same as $\displaystyle ln(\frac{3}{1}) = ln(3)$

correct or have I misunderstood something?

6. Originally Posted by wizzler
oh right.. so then i can just put in 3 and 1 for the function because if x = 1 then t = 3; if x = -1 then t = 1 so I get: (Note it's a definite integral!!!)

$\displaystyle ln(3) - ln(1)$ and that is the same as $\displaystyle ln(\frac{3}{1}) = ln(3)$

correct or have I misunderstood something?
Yup.

-Dan