# Slope of this tangent

Okay I have to find the slope of this tangent to the curve $y=\int_0^x \frac{dx}{1+x^3}$ at the point where $x=1$.

My try- I integrated the expression and differentiated it afterwards to get the slope. But I'm not getting the correct answer on putting $x=1$ in the final expression that I'm getting after integration and diffrentiation.

• Are you sure you don't mean $$y = \int_0^x \frac{dt}{1+t^3}$$ – ÍgjøgnumMeg Jul 9 '16 at 9:55
• you want to calculate slope of the tangent than its of form $y=mx+c$ where m is the slope – Nebo Alex Jul 9 '16 at 9:56
• I'm simply pointing out what @okrzysik said; $t$ is a dummy variable. – ÍgjøgnumMeg Jul 9 '16 at 10:02
$$\frac{dy}{dx} = \frac{1}{1+x^3} \Rightarrow \frac{dy}{dx}\bigg|_{x=1} = \frac{1}{1+1^3} = \frac{1}{2}$$