# slope of a tangent line through point [duplicate]

I was given this calculus problem and am very stuck on it and cant get an answer

The slope of the tangent line to the graph of $f(x)$ at each $x\neq 0$ is given by $e^{6x}+\frac{6}{x}$ and knowing that the graph contains the point $(1,e/3)$, find $f(x)$.

I know you need to integrate but I am stuck and cant get the right answer can anyone walk me through it?

• Formatting tips here. – Em. Jun 16 '16 at 6:31
• What is this ? You are asking the same question YOU YOURSELF ASKED YESTERDAY? And YOU had got answers as well!! – Qwerty Jun 16 '16 at 6:35
• You're supposed to ask respondents for clarification, not make another post. – Em. Jun 16 '16 at 6:39

You know that $f(1)=\frac{e}{3}$ and that $\forall x \geq 1,f'(x)=e^{6x}+\frac{6}{x}$.
So $$\forall x \geq 1,f(x)=\int_1^xf'(t)dt+f(1)=\int_1^xe^{6t}dt+\int_1^x\frac{6}{t}dt+\frac{e}{3}=\frac{1}{6}(e^{6x}-e^6)+6\ln(x)+\frac{e}{3}$$
Depending of the domain of $f$ you can find the expression of $f$ for other values of $x$ by using the same idea.