# Evaluating Definite Integral

1. Jan 26, 2005

If I want to evaluate: $$\int^1_0 \frac {dx}{(x+1)^2}$$ I need to use the Fundamental Theorem of Calculus right? SO wouldn't I have to solve $$\int^b_a \frac{dx}{(x+1)^2} = \frac{(x+1)^3}{3} = \frac {8}{3} - \frac {1}{3}$$? But the answer is $$\frac {1}{2}$$

Last edited: Jan 26, 2005
2. Jan 26, 2005

### dextercioby

There are a lot of errors there.How about making a substitution ??

Daniel.

P.S.You evaluated wrongly another integral,not the one u were supposed to...

3. Jan 26, 2005

### kingyof2thejring

you're gona integrate that function.
between the limits 1 and 0
1/(x+1)^2 is the same as (x+1)^-2 {to the power neg two}
when you integrate you add 1 to the power

(x+1)^(-2+1) {to the power neg one}
and then divide by this new value of the power
(x+1)^-1
-1
rearanging the equation gives -1/(x+1)
when you sub 1 for x you get -1/2
when you sub 0 for x you get 1
-1/2 + 1 you get 1/2

4. Jan 26, 2005

I got it. It's $$\int^1_0 \frac {dx}{(x+1)^2} = \frac {-1}{x+1}$$
So $$F(1) - F(0) = \frac {1}{2}$$

5. Jan 26, 2005

### dextercioby

Not really...What about the integration limits?

As for the limit part,could u rewrite it in an intelligible form...?

Daniel.

6. Jan 26, 2005

I am not sure what you mean. I was just using the fact that $$\int^b_a f(u) \ du = F(b) - F(a)$$. Why wouldn't my answer be correct?

Thanks

7. Jan 26, 2005

### dextercioby

Because it shouldn't depend on "x"...It should be a real number...Not a function...

Daniel.

8. Jan 26, 2005

### dextercioby

Where did the rest of the limit go??

Daniel.