= Solution
The vertical integral is <absolutely convergent> because the denominator is quadratic in the imaginary part. If $x<1$, move its line to the right: no poles are crossed, and the integral on $\operatorname{Re}s=R$ is $O(x^R/R)$, tending to <zero>. If $x>1$, move left across the simple poles at $0,-1$. Their residues are $1$ and $-x^{-1}$, and the far-left integral tends to <zero>. Horizontal segments vanish by their $O(|\operatorname{Im}s|^{-2})$ decay for a fixed pair of vertical lines. At $x=1$, dominated convergence on the original line gives the limiting value <zero>. Hence the <kernel> is $0$ below one and $1-x^{-1}$ at or above one.
On a line $c>2$, <absolute convergence> permits interchanging the totient <Dirichlet series> and this integral. The $n$th term is $x\phi(n)$ times the <kernel> at $x/n$, giving exactly $\phi(n)(x-n)$ for $n\le x$. Thus
$$
S(x):=\sum_{n\le x}\phi(n)(x-n)
=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}
\frac{\zeta(s-1)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}\,ds.
$$
We may take $c=5/2$ and shift to $\operatorname{Re}s=3/2$. The denominator has no zeros in this strip; the only <pole> crossed is $s=2$, where $\zeta(s-1)$ has <residue> one. Its <residue> in the integrand is
$$
\frac{x^3}{6\zeta(2)}=\frac{x^3}{\pi^2}.
$$
On the new line, the supplied estimates bound the integrand by $O(x^{5/2}|t|^{-3/2}\log^8|t|)$ for $|t|\ge2$, an integrable function. The compact portion is bounded as well. The same estimates, uniformly across the strip, make the horizontal integrals tend to <zero>. Consequently
$$
\boxed{S(x)=\frac{x^3}{\pi^2}+O(x^{5/2})\sim\frac{x^3}{\pi^2}.}
$$
For the Tauberian step let $A(x)=\sum_{n\le x}\phi(n)$. It is nondecreasing because the summands are nonnegative, and $S(x)=\int_0^xA(t)\,dt$. For fixed $0<\epsilon<1$,
$$
\frac{S(x)-S((1-\epsilon)x)}{\epsilon x}
\le A(x)\le
\frac{S((1+\epsilon)x)-S(x)}{\epsilon x}.
$$
After division by $x^2$ and passage to lower and upper limits, the asymptotic for $S$ gives bounds $(3-3\epsilon+\epsilon^2)/\pi^2$ and $(3+3\epsilon+\epsilon^2)/\pi^2$. Letting $\epsilon\downarrow0$ squeezes them together. Therefore
$$
\boxed{\sum_{n\le x}\phi(n)\sim\frac{3x^2}{\pi^2}.}
$$
This uses monotonicity of the summatory function; differentiating an asymptotic formula without such control would not justify the conclusion.
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