f'(x) = (x*cosx + 7) /x

Let us simplify the function.

f'(x) = xcosx/x + 7/x

= cosx + 7/x

==> f'(x) = cosx + 7/x

Now we know that f(x) = integral f'(x).

==> f(x) = intg f'(x) dx

= intg (cosx + 7/x) dx

= intg cosx dx + intg 7/x dx

= sinx + 7*ln x + C

**==>f(x) = sinx + 7*lnx + C**

**f(0) is not a valid point because lnx is not defined at x= 0.**

We are given that f'(x) = ( x*cos x + 7) /x and f(0) = 5.

Now f(x) = Int (f'(x))

Int [f'(x)] = Int [( x*cos x + 7) /x]

=> Int [ cos x + 7/x] +C

=> Int (cos x) + Int (7/x) + C

=> sin x + 7 ln x + C

Now, f(0) = 5

=> sin 0 + 7 ln 0 + C = 5

But ln 0 is not defined.

**Therefore f(x) = sin x + 7 ln x + C.**

f'(x) = (x*cosx+7)/x and f'(0) = 5. To find f(x).

Since f'(x) = (x*cosx+7)/x,

f(x) = Int(x*cosx+7)/x dx = Int(cosx+7/x) dx = -sinx +7lnx+C

f'(0) = 5 given

So f(x) = -sinx +7lnx +C.

Since f(0) = Lt x--> 0 ln(x) = -infinity. So f(0) does not exists.

So f(x) = -cosx+7lnx+c.