Originally Posted by
Chris L T521 There way is correct, and so is mine.
When we integrate $\displaystyle \frac{1}{(x+1)^2}$, we still need to apply a substitution.
The integral would become $\displaystyle \int\frac{\,du}{u^2}=-\frac{1}{u}$
What they did was re-substitute u back into the antiderivative:
$\displaystyle -\frac{1}{u}=-\frac{1}{x+1}$, and then they evaluated from 1 to 2.
However, the way I did it, I eliminated the need to re-substitute u back into the equation.
I did this when I converted the integration limits.
Since $\displaystyle u=x+1$ and the x limits of integration were from 1 to 2, I saw that the u limits would be from $\displaystyle u=1+1=\color{red}\bold 2$ to $\displaystyle u=2+1=\color{red}\bold 3$
So when you evaluate, we see that
$\displaystyle \int_1^2\frac{\,dx}{(x+1)^2}=\int_2^3\frac{\,du}{u ^2}=\color{red}\boxed{\frac{1}{6}}$
I hope this clarified things! If you have questions, feel free to ask. We're here to help you!
--Chris