In a course "mathematical physics" we just encountered distribution, we covered the topic distribution in about 4 hours, after the lectures i think i understood it but when i look at the exercises i realise i don't understand anything at all...
Consider a test function $\phi \in C^{\infty}_{[0,1]}$ and a function $f$ continous in $[0,1].$ The plus distribution $[\frac{f(x)}{1-x}]_{+}$ is defined as
$$T_{[\frac{f(x)}{1-x}]_{+}}[\phi] = \int^1_0 dx\frac{f(x)}{1-x}(\phi(x)-\phi(1))$$
which implies $$[\frac{f(x)}{1-x}]_{+}=\frac{f(x)}{1-x}-\delta(x-1)\int^1_0dt\frac{f(t)}{1-t}$$
Now i'm supposed to compute the integral $$I=\int^1_0dx\frac{x^2+3}{(1-x)_{+}}$$
I think we can say that we can rewrite the whole thing:
$$I=\int^1_0dx\frac{x^2+3}{(1-x)_{+}}=\frac{x^2+3}{1-x}_{[+]}=\frac{x^2+3}{1-x}-\delta(x-1)\int^1_0dt\frac{t^2+3}{1-t}$$
But then I don't know what to do about $\delta(x-1)$. And this just feels weird. Can someone give me a hand?
EDIT: Based on the comment of @JeanMarie, I try again
\begin{align} I & = \int^1_0dx\frac{x^2+3}{(1-x)_{+}} \\ & = \frac{x^2+3}{1-x}_{[+]} \\ & = \frac{x^2+3}{1-x}-\delta(x-1)\int^1_0dt\frac{t^2+3}{1-t} \\ & = (\frac{(1-x)^2}{1-x}-\frac{2(1-x)}{1-x})-\int^1_0 \delta(x-1)\frac{(1-t)^2}{1-t}-\int^1_0 \delta(x-1)\frac{2(1-t}{1-t} \\ & =-(x+1)-\int^1_0 \delta(x-1)(1-t)-\int^1_0 \delta(x-1)2 \end{align}
That's what I've got so far, i know how to evaluate those integrals with delta function inside, but they have different variables, what should one do about that?