Solve $f$ by $f(x)=17+\int_0^x\frac{f(t)}{(t+2)(t+3)}\,dt$ Solve $f(1)$ by $f(x)=17+\int_0^x\frac{f(t)}{(t+2)(t+3)}\,dt$ when $x>0$.
My attempt was to differentiate both side and get $f'(x)=\frac{f(x)}{(x+2)(x+3)}$. However, I can't continue going on. Need help.
 A: So $${y'\over y} = {1\over x+2}-{1\over x+3}$$ and thus $$\int {dy\over y} = \int ({1\over x+2}-{1\over x+3})dx$$ which give us: $$\ln y = \ln{x+2\over x+3}+c\implies y=A\cdot {x+2\over x+3}$$
Note that $f(0)= 17$ so $A=51/2$ and thus $f(1) = {3\over 4}\cdot {51\over 2}= ...$
A: It follows from what you wrote that\begin{align}(\log\circ f)'(x)&=\frac1{(x+2)(x+3)}\\&=\frac1{x+2}-\frac1{x+3}\\&=\log'(x+2)-\log'(x+3).\end{align}Can you take it from here?
A: $$f(x)=17+\int_{0}^{1} \frac{f(t)}{(t+2)(t+3)}dt $$
D.w.r.t $x$, we get
$$\frac{f'(x)}{f(x)}=\frac{1}{(x+2)(x+3)}=\frac{1}{(x+2)}-\frac{1}{(x+3)}$$
Integrating we have,
$$f(x)=C\frac{x+2}{x+3}, ~~C=51/2$$
A: Use the first part of the Fundamental Theorem of Calculus. From the first part
$$ f(x) = 17 + \int_0^x \frac{f(t)}{(t + 2)(t + 3)}dt $$
differentiating both sides give
$$ f’(x) = \frac{f(x)}{(x + 2)(x + 3)} $$
divide both sides by $f(x)$
$$ \frac{f’(x)}{f(x)} = \frac{1}{(x + 2)(x + 3)}$$
Partial Fraction Decomposition gives the RHS
$$ \frac{A}{x + 2} + \frac{B}{x + 3} $$
So when $x = -2$, $A = 1$ and when $x = -3$, $B = -1$ and therefore, RHS would be
$$ \frac{f’(x)}{f(x)} = \frac{1}{x + 2} - \frac{1}{x + 3} $$
Now, Integrate both sides with respect to x. Your equation now becomes
$$ \int\frac{f’(x)}{f(x)}dx = \int \frac{1}{x + 2} - \frac{1}{x + 3} dx $$
$$ = ln |f(x)| = ln |x + 2| - ln |x + 3| + C $$
Clean things up
$$ ln|f(x)| = ln\left|\frac{x + 2}{x + 3}\right| + C $$
exponentiating both sides
$$ e^{f(x)} = e^{ln\left|\frac{x + 2}{x + 3}\right| + C} $$
$$ f(x) = \frac{x + 2}{x + 3} C $$
We can say $e^C$ is also equal to $C$ because it’s the same as the constant of integration. Now let us try finding $f(0)$ by relating $f(x)$ into our current equation. In our original equation;
$$f(x) = 17 + \int_0^x \frac{f(t)}{(t + 2)(t + 3)}dt$$
$$\frac{(x + 2)}{(x + 3)}C = \int_0^x\frac{f(t)}{(t + 2)(t + 3)}dt$$
So when $x = 0$ we are left with $17$ alone in the right side of our equation because any integral that contains both limits equal to one another evaluates straightly to zero. 
$$f(0) = 17$$
$$\frac{2}{3}C = 17$$
$$C = 17\left(\frac{3}{2}\right)$$
$$C = \frac{51}{2}$$
So, when x = 1
$$ f(1) = \frac{3}{4} C$$
$$ f(1) = \left(\frac{3}{4}\right) \left(\frac{51}{2}\right)$$
$$ f(1) = \frac{153}{8}$$
