How Long to Reach Speed of Light?

[JakePearson]

if a spaceship accelerates, from restat time t = 0, at a rate of 2t / sqrt(1 + t^2) m/s at time t, calculate in years to 1 significant figure how long it would take to reach the speed of light?

my attempt

speed of light = 3.0 x 10^8 m/s
integrate the rate function 2t / sqrt(1 + t^2) from t = 0, to the t we are looking for

integrating 2t / sqrt(1 + t^2)

using integration by parts;

let u = 1 + t^2, then du = 2t dt

WHERE DO I GO FROM HERE

 

[Mark44]

Before you go at an integral using integration by parts, you should always see if a simpler substitution will work. What you have shown is exactly the substitution I would use (your work does not show that you are doing integration by parts).

Using this substitution, what does your new integral look like?

 

[HallsofIvy]

if a spaceship accelerates, from restat time t = 0, at a rate of 2t / sqrt(1 + t^2) m/s at time t, calculate in years to 1 significant figure how long it would take to reach the speed of light?

my attempt

speed of light = 3.0 x 10^8 m/s
integrate the rate function 2t / sqrt(1 + t^2) from t = 0, to the t we are looking for

integrating 2t / sqrt(1 + t^2)

using integration by parts;

let u = 1 + t^2, then du = 2t dt

WHERE DO I GO FROM HERE

That is NOT "integration by parts". Perhaps you should review that.
You want to integrate
$$\int \frac{2t dt}{\sqrt{1+ t^2}}$$
and you say $u= 1+ t^2$ and $du= 2tdt$. Okay, doesn't it make sense to replace the "2tdt" in the integral by du and the "$1+ t^2$ in the integral by u?
You might want to remember that $1/\sqrt{a}= a^{-1/2}$.

 

[JakePearson]

cheers, my mistake