Area under the graph and Integrals
Solution 1:
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.