How to find inverse of convolution integral? If $x^{-1}(t)$ and $ y^{-1}(t)$ denote the integrals of x(t) and y(t)  defined by

$x^{-1}(t)=\displaystyle\int_{-\infty}^{t} x(\lambda)d\lambda$

$y^{-1}(t)=\displaystyle\int_{-\infty}^{t}y(\lambda)d\lambda$

then,

$(x*y)^{-1}$ which denotes the inverse integral of (x*y )is equal to:

(a) $x^{-1}*y^{-1}$

(b) $x*y$

(c)$x^{-1}*y$

(d)$x*y^{-1}$

i tried it follows:

from both conditions above we can easily infer following

$\dfrac{d}{dt}x^{-1}(t)=x(t)\implies sX^{-1}(s)=X(s)$

$\dfrac{d}{dt}y^{-1}(t)=y(t)\implies sY^{-1}(s)=Y(s)$

then i multiplied both(keeping in mind multiplication in freq. domain corresponds to convolution in time domain)

$s^2[X^{-1}(s)Y^{-1}(s)]=X(s).Y(s)$

$s[X^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[sX^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}] \implies x*y^{-1}=(x*y)^{-1}$

and also,

if i would've written as

$s[X^{-1}(s)Y^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X^{-1}(s)sY^{-1}(s)]=\dfrac{X(s).Y(s)}{s}$

$[X^{-1}(s)Y(s)]=\dfrac{X(s).Y(s)}{s}\implies x^{-1}*y=(x*y)^{-1}$

so, options (c,d) both should be correct but in the book the answer is given as

only option (c)

so , where i am doing mistake, and what should be the correct answer, i'm stuck

with this book's faulty answers...so,please help
The integral of a signal $x(t)$

$$f(t)=\int_{-\infty}^tx(\tau)d\tau$$

can be written as the convolution of $x(t)$ with the unit step function $u(t)$:

$$f(t)=(x\star u)(t)$$

Consequently, the integral of a convolution

$$g(t)=\int_{-\infty}^t(x\star y)(\tau)d\tau$$

can be written as

$$g(t)=((x\star y)\star u)(t)$$

And due to associativity of convolution we have

$$g(t)=((x\star u)\star y)(t)=((y\star u)\star x)(t)$$

So options $(c)$ and $(d)$ in your question are both correct.