Solve $\int_{1}^{x} \frac{dt}{f'(t)+3} = \frac{1}{x} + x^2 -2$ for $f(x)$ 
Solve $$\int_{1}^{x} \frac{dt}{f'(t)+3} = \frac{1}{x} + x^2 -2$$ for $f(x)$.

I think I should use the fundamental theorem of calculus but I'm not sure how. Can I say that $$\int_{1}^{x} \frac{dt}{f'(t)+3} = \frac{1}{f(t)+3}?$$
I tried to integrate the other part as follows $$\int\left(\frac{1}{x} + x^2 -2\right)dx = \ln(x) + \frac{x^3}{3} -2x + k,\,k\in \mathbb{R}$$ but then I don't know what to do.
The answer is $f(x)=\frac{1}{6}\ln(2x^3-1)-3x+4$.
 A: Use Leibniz's rule to differentiate (not integrate) both sides. This gives $\displaystyle\frac1{f'(x)+3}=-\frac1{x^2}+2x$ so $\displaystyle f'(x)=\frac{x^2}{2x^3-1}-3$ which becomes a very straightforward substitution integral.
A: Hint:  This is an integral equation of the form $\int_a^x g(f'(t))dt=h(x)$, where you need to solve for $f(x)$, $g(x)$ is in a form that is easily inverted where $g^{-1}(y(x))$ is easily integrable if $y(x)$ is easily integrable, and $h(x)$ is an arbitrary differentiable function fulfilling $h(a)=0$, try taking the following steps:

*

*Differentiate both sides with respect to $x$ and get $g(f'(x))=h'(x)$

*Apply $g^{-1}$ to both sides and get $f'(x)=g^{-1}(h'(x))$

*Integrate both sides with respect to $x$
In this particular instance, $a=1$, $h(x)=x^{-1}+x^2-2$, which is differentiable and fulfills $h(1)=0$, and $g(x)=\frac{1}{x+3}$.  When you get up to integration, try substituting $u=2x^3-3$
A: let $g(t)=\frac{1}{f'(t)+3}$ and $G'=g$ then we have:
$$G(x)-G(1)=\frac1x+x^2-2$$
now differentiate both sides and we get:
$$g(x)=-\frac1{x^2}+2x$$
sub back in:
$$\frac{1}{f'(x)+3}=2x-\frac1{x^2}$$
rearrange:
$$f'(x)=\frac{x^2}{2x^3-1}-3$$
and now finally:
$$f(x)=\int f'(x)dx$$
