Question: > 
> Find the area under the graph of
> $$
> f(t) = \frac{t}{(1+t^2)^a}
> $$
> between $t=0$ and $t=x$ where $a>0$ and $a \neq 1$ is fixed, and evaluate the limit as $x \to \infty$.
> 
> 
> 

Hello, for this question I tried using substitution but the different conditions like $t=0$ and $t=x$ confuse me. Especially what "$a$" is, since $a$ is positive and not equal $1$, I'm completely stuck there. Please explain how I would approach this question. I know substitution is neccesarly

Answer: 
Hint: substitute $1 + t^2 = u$. If $t = 0$, what is $u$? If $t = x$, what is $u$?

> 
>  \begin{align} 1 + t^2 = u &\Rightarrow 2tdt = du\\ t = 0 &\Rightarrow u = 1+0^2 = 1\\ t = x &\Rightarrow u = 1+x^2 \end{align}
> 
> 
> 

$a$ is a parameter, which you should consider as a fixed number.
You are correct that a substitution will work here. Namely, $u = 1+t^2$, which has differential $du = 2t \, dt$ and $u = 1 + 0^2 = 1$ and $u = 1 + x^2$. Can you make the substitution and complete the integral? (Treat $a$ and $x$ as if they are numbers.) Click to reveal:

> 
>  \begin{align} \int\_0^x \!\! \frac{t}{(1+t^2)^a} \, dt &= \frac12 \int\_1^{1+x^2} \!\! u^{-a} \, du \\ &= \biggl.\frac{1}{2(1-a)} \cdot \frac{1}{u^{a+1}} \biggr\rvert\_1^{1+x^2} \\ &= \frac{1}{2(1-a)} \biggl( \frac{1}{(1+x^2)^{a+1}} - 1 \biggr) \\ &= \frac{1}{2(a-1)} \biggl( 1 - \frac{1}{(1+x^2)^{a+1}} \biggr) \end{align}
> 
> 
> 

This expression gives the value of the definite integral in terms of the quantities $a$ and $x$. Now, if you want to let $x$ increase without bound, you evaluate the the improper integral as a limit:

> 
>  \begin{align} \int\_0^\infty \!\! \frac{t}{(1+t^2)^a} \, dt &= \lim\_{x \to \infty} \int\_0^x \!\! \frac{t}{(1+t^2)^a} \, dt \\ &= \lim\_{x \to \infty} \frac{1}{2(a-1)} \biggl( 1 - \frac{1}{(1+x^2)^{a+1}} \biggr) \\ &= \frac{1}{2(a-1)}. \end{align}
> 
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