How to use dimensional analysis to find these integrals? 
Use dimensional analysis to find $\displaystyle \int_0^∞ \mathrm{e}^{-ax} \,\mathrm{d}x$ and $\displaystyle \int\frac{\mathrm{d}x}{x^2 + a^2}$.  A useful result is$$
\int \frac{\mathrm{d}x}{x^2 + 1} = \arctan x + C.
$$

I am using the the book called street mathematics to learn more about dimensional analysis. I am trying to understand a problem in the book. The question is to use dimensional analysis to find the solutions to two integrals. both are in the attached photo.
I tried to understand the question and how to best tackle it but I did not succeed.
 A: Start by giving simple dimensions to $x$, say $[x] = L$, length (the author uses $[x]$ to denote the dimension of $x$). Functions like the exponential and sine/cosine take dimensionless quantities as input and give dimensionless quantities as output. This means $[a][x]=1$ (dimensionless), or $[a] = \dfrac{1}{L}$, $[e^{-ax}]=1$ (dimensionless), and $[dx]=L$.
Since $a$ is a parameter in $\displaystyle \int_0^{\infty} e^{-ax}dx$, the value of the integral depends on $a$. The integral has dimensions of $L$, which leads us to try to build an answer out of $a$ with those dimensions. The only choice with the correct dimensions is $\dfrac{1}{a}$:
$$\int_0^{\infty} e^{-ax}dx =\dfrac{C}{a}$$
where $C$ is a dimensionless number.
A: Calling this "dimensional analysis" is not accurate. I believe the idea behind this is if you change the "dimension" of the variable, it multiplies the value by a constant. If you choose the constant wisely, then it simplifies the integral to one that is known (which is why he gives you one of the simplified integrals - I assume the other integral was already covered).
However, this is just a very limited form of what is a well-known and simple trick: substition.
In the first integral, we change the units on $x$ to exactly balance the $a$ - that is, $ax$ becomes $x$ in the new units. To make this a little clearer, lets call the variable in the new unit $u$. So $u = ax$. Assuming that $a > 0$ (if it isn't, the integral doesn't converge), When $x = 0, u = a(0) = 0$, and when $x = \infty, u = a(\infty) = \infty$. Now small changes in $x$ induce small changes in $u$ that are $a$ times as big: $du = a\,dx$. So the integral becomes
$$\int_0^\infty e^{-ax}\,dx = \int_0^\infty e^{-u}\,\frac{du}a = \frac 1a \int_0^\infty e^{-u}\, du =\frac 1a (1) = \frac 1a$$
For the second, if we pull $a^2$ out of the top and bottom of the fraction, we get $$\int \dfrac{\frac{dx}{a^2}}{\frac{x^2}{a^2} + 1} = \int\dfrac{\frac 1ad\left(\frac xa\right)}{\left(\frac xa \right)^2 + 1}$$
I'll leave the rest to you.
