### Video Transcript

Determine the function π of π‘
such that π prime of π‘ is equal to negative two divided by three times π‘ squared
plus one and π evaluated at one is equal to zero.

The question wants us to determine
the function π of π‘. The information weβre given is that
the derivative of π of π‘ with respect to π‘ is equal to negative two divided by
three times π‘ squared plus one and π evaluated at one is equal to zero. This is a differential
equation. In fact, itβs a simple differential
equation because weβre just given π prime of π‘ in terms of π‘. So to solve this, weβre going to
integrate both sides of this equation with respect to π‘. This will give us our function π
of π‘ up to a constant of integration. This gives us the integral of π
prime of π‘ with respect to π‘ is equal to the integral of negative two divided by
three times π‘ squared plus one with respect to π‘.

Remember, integration and
differentiation are opposite processes. So, the integral of π prime of π‘
with respect to π‘ is π of π‘ up to a constant of integration. So to find π of π‘, we just need
to evaluate our integral. Before we evaluate our integral,
weβll take out our constant factor of negative two divided by three. This gives us negative two over
three times the integral of one divided by π‘ squared plus one with respect to
π‘. But this integral is just one of
our standard integrals involving inverse trigonometric functions.

We know the integral of one divided
by π₯ squared plus one with respect to π₯ is equal to the inverse tan of π₯ plus the
constant of integration πΆ. So, we can just apply this to
evaluate our integral. We get negative two over three
times the inverse tan of π‘ plus πΆ. So, this is the general solution to
our differential equation. Remember, we want the solution
where π evaluated at one is equal to zero. So, weβll substitute this
information into our general solution to find the value of πΆ.

This gives us zero is equal to
negative two-thirds times the inverse tan of one plus πΆ. We can simplify this equation. We can divide both sides through by
negative two-thirds. Next, we can evaluate the inverse
tan of one; itβs equal to π by four. Finally, we subtract π by four
from both sides of this equation. We see that πΆ is equal to negative
π by four. All we need to do now is use this
value of πΆ in our general solution.

Doing this, we get π of π‘ is
equal to negative two over three times the inverse tan of π‘ minus π over four. And we could leave our answer like
this. However, weβll distribute negative
two-thirds over our parentheses. This gives us negative two times
the inverse tan of π‘ divided by three plus π by six. And this gives us our final
answer.

Therefore, weβve shown if π prime
of π‘ is equal to negative two over three times π‘ squared plus one and π of one is
equal to zero. Then π of π‘ is equal to negative
two times the inverse tan of π‘ divided by three minus π by six.