The derivation of the formula for variance for a Pareto Distribution

You indeed have the right ideas, but probably missing a trick in computing the integrals. Below is how I would do it. 

First note that if $X \sim \operatorname{Pareto}(a,\lambda)$ with the density function $$f_X(x) = \frac{a \lambda^a}{x^{a+1}}, \quad x \ge \lambda,$$ we have $$\int_{x=\lambda}^\infty f_X(x) \, dx = a \lambda^a \int_{x=\lambda}^\infty x^{-(a+1)} \, dx = a\lambda^a \left[-\frac{1}{a x^a} \right]_{x = \lambda}^\infty = 1.$$ Hence the density integrates to $1$ for any $a > 0$.

Now consider the $k^{\rm th}$ raw moment, which is given by

$$\operatorname{E}[X^k] = \int_{x=\lambda}^\infty x^k \cdot \frac{a \lambda^a}{x^{a+1}} dx = a \lambda^a \int_{x=\lambda}^\infty \frac{1}{x^{a+1-k}} dx$$
$$ = \frac{a \lambda^a}{(a-k)\lambda^{a-k}} \int_{x=\lambda}^\infty \frac{(a-k) \lambda^{a-k}}{x^{a-k+1}} dx$$
$$= \frac{a\lambda^k}{(a-k)}\int_{x=\lambda}^\infty \frac{a' \lambda^{a'}}{x^{a'+1}} dx= \frac{a\lambda^k}{(a-k)}, $$
where $a'=a-k$. This is because the last integral is the integral of a $\operatorname{Pareto}(a', \lambda)$ density function, and hence it equals $1$ whenever $a'=a-k > 0$. In particular we have $$\operatorname{E}[X] = \frac{a\lambda}{a-1}, \quad \operatorname{E}[X^2] = \frac{a\lambda^2}{a-2},$$ corresponding to the choices $k = 1$ and $k = 2$, respectively.

Now for or $a > 1$, and the second for $a > 2$, the variance is given by
$$ \operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \frac{a\lambda^2}{a-2} - \frac{a^2 \lambda^2}{(a-1)^2}$$
$$= \frac{((a-1)^2 - a(a-2))a\lambda^2}{(a-1)^2(a-2)} = \frac{a\lambda^2}{(a-1)^2(a-2)}.$$