Added: I interpreted the question as follows: suppose that we use the integration by parts formula,
$$\int u(x)v'(x)\,dx = u(x)v(x) - \int v(x)u'(x)\,dx$$
and $\int v(x)u'(x)\,dx = \alpha \int u(x)v'(x)\,dx$ for some constant $\alpha$, so that we can "solve" for the original integral as
$$\int u(x)v'(x)\,dx = \frac{1}{1+\alpha}u(x)v(x) + C.$$
Is it the case that we can solve the original integral by doing a direct substitution instead of integration by parts?
I think this works:
Suppose that $v(x)u'(x) = \alpha u(x)v'(x)$ for some $\alpha\neq -1$. (I think this is essentially what you have, since the integral on the left hand side is $\int u(x)v'(x)\,dx$, and the integral on the right is $\int v(x)u'(x)\,dx$). I restrict to $\alpha\neq -1$, because if $\alpha=-1$, then you cannot "solve" for the original integral. But in fact, from the work below we will get that $u=\frac{B}{|v|}$, so that integration by parts will simply result in the true (but useless) $\int u\,dv = B + \int u\,dv$.
If $u$ or $v$ are zero, then the original integral was the integral of $0$, so we may discard that case. So we get that $\frac{u'}{u} = \alpha \frac{v'}{v}$, and integrating we get $\ln|u|=\alpha\ln|v|+C$, or $|u| = A|v|^{\alpha}$ for some $A\gt 0$; hence $u=B|v|^{\alpha}$ for some $B\neq 0$. So the original integral was
$$\int u(x)v'(x)\,dx = \int B|v(x)|^{\alpha}v'(x)\,dx$$
which suggests the substitution $g=v(x)$ to get $B\int |g|^{\alpha}\,dg$.
Added: Note that if $\alpha = -1$, we still get $u=B|v|^{-1}$, and
so again the substitution $g=|v(x)|$ will do the trick.
Only one application is necessary; letting $u = 3x$ and $dv = \cos 2x dx$, so that $du = 3 dx$ and $v = \frac 1 2 \sin 2x$, we find that
$$\int 3x \cos(2x) dx = \frac{3x}{2} \sin 2x - \frac 3 2 \int \sin 2x dx$$
Now can you find the antiderivative of $\sin 2x$?
Best Answer
Take $u=e^{x^2}$ then $du=2xe^{x^2}$ and $v=\frac{x^2}{2}$ so $$\int xe^{x^2}dx=\frac{x^2e^{x^2}}{2}-\int x^3e^{x^2}dx$$ Take $u=e^{x^2}$ and $dv=x^3$ then $$\int x^3e^{x^2}dx=\frac{x^4}{4}e^{x^2}-\frac{1}{2}\int x^5e^{x^2}dx$$ Similary $$\int x^5e^{x^2}dx=\frac{x^6}{6}e^{x^2}-\frac{1}{3}\int x^7e^{x^2}dx$$ In general $$\int x^{2n+1}e^{x^2}dx=\frac{x^{2n+2}}{2n+2}e^{x^2}-\frac{1}{n+1}\int x^{2n+3}e^{x^2}dx$$ From that we can take a few terms and notice that $$\int xe^{x^2}dx=\frac{x^2}{2}e^{x^2}-\frac{x^4}{4}e^{x^2}+\frac{x^6}{12}e^{x^2}-\frac{x^8}{48}e^{x^2}+\frac{x^{10}}{240}e^{x^2}+\cdots\\\int xe^{x^2}dx=\frac{e^{x^2}}{2}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k}}{k!}$$Now with few adjustments you'll get$$e^{-x^2}=\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{k!}=1+\sum_{k=1}^\infty\frac{(-1)^kx^{2k}}{k!}=1-\sum_{k=1}^\infty\frac{(-1)^{k+1}x^{2k}}{k!}\\\int xe^{x^2}dx=\frac{e^{x^2}}{2}\left(1-e^{-x^2}\right)=\frac{e^{x^2}}{2}+C$$