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 could use the reverse-chain rule:

Let $u = 1 + t^2$,

Then you have $du = 2t 
\times dt$

Hence, your integral becomes:

$$\int{\frac{1}{u^a} \times t}{dt} \to \frac{1}{2}\int{\frac{1}{u^a} \times 2t}{dt} \to \frac{1}{2}\int{\frac{1}{u^a} du}$$

Since $a ≠ \{0, 1\}$ and $a > 0$ you can apply the index laws in integration.

$$\frac{1}{2}\int{\frac{1}{u^a} du} = \frac{1}{2}\int{u^{-a} du} = \frac{1}{2} (\frac{1} {-a + 1} \times u^{-a + 1}) + C$$