### Video Transcript

The slope of the tangent to the curve π¦ is equal to π of π₯ at the point π₯, π¦ is given by three π₯π to the two π₯ divided by two π₯ plus one squared. Determine the function π of π₯ if the point one, five π squared lies on the curve.

The question tells us the slope to the curve π¦ is equal to π of π₯ at the point π₯, π¦ is given by three π₯π to the power of two π₯ all divided by two π₯ plus one squared. In other words, weβre told π prime of π₯ is equal to three π₯π to the two π₯ divided by two π₯ plus one squared. The question wants us to determine the function π of π₯ given that the point one, five π squared lies on our curve π¦ is equal to π of π₯.

To do this, weβre going to notice that π of π₯ is an antiderivative of its slope function π prime of π₯. In other words, our function π of π₯ is equal to the integral of π prime of π₯ with respect to π₯ up to our constant of integration π. So, we need to integrate three π₯π to the power of two π₯ divided by two π₯ plus one squared with respect to π₯. We notice that our integrand is not in a standard form which we can integrate. So, weβre going to need to perform some kind of manipulation.

We could try using the substitution π’ is equal to two π₯ plus one. However, this does not make current to go any easier. So, weβre going to try using integration by parts. We recall integration by parts tells us the integral of π’π£ prime with respect to π₯ is equal to π’π£ minus the integral of π’ prime π£ with respect to π₯. It might be tempting at this point to try and set our function π’ to be some kind of algebraic function. However, weβre going to set our function π’ to be the numerator of our fraction, three π₯ multiplied by π to the power of two π₯.

Weβll also write π£ prime to be one divided by the denominator of our fraction. Weβll write this as two π₯ plus one all raised to the power of negative two. To calculate π’ prime, weβre going to use the product rule. First, we differentiate three π₯ to get three. Then, we multiply this by π to the power of two π₯. Then, we add on the derivative of π to the power of two π₯. Thatβs two π to of the power of two π₯. And we multiply this by three π₯.

Now, we can notice that both terms in our expression for π’ prime share a factor of three π to the power of two π₯. So, we can factor this out, giving us three π to the two π₯ multiplied by one plus two π₯. To find our antiderivative of the function π£ prime, weβre going to use integration by substitution. Weβll use the substitution π€ is equal to two π₯ plus one. Differentiating both sides of this equation with respect to π₯ gives us that the derivative of π€ with respect to π₯ is equal to two.

And although dπ€ by dπ₯ is not a fraction, when using integration by substitution, it behaves a little bit like one. This gives us the equivalent statement that half dπ€ is equal to dπ₯. So, by using the substitution π€ is equal to two π₯ plus one, we have that π£ is equal to the integral of π€ raised to the power of negative two multiplied by a half with respect to π€. We can integrate this by adding one to our exponent and then dividing by the new exponent. This gives us that π£ is equal to negative π€ to the power of negative one divided by two.

Finally, by substituting π€ is equal to two π₯ plus one and bringing π€ down into our denominator, we have that π£ is equal to negative one divided by two multiplied by two π₯ plus one. So, by using integration by parts, weβve shown that our function π of π₯ is equal to π’ multiplied by π£, which is three π₯π to the power of two π₯ multiplied by negative one divided by two multiplied by two π₯ plus one. And then, we subtract the integral of π’ prime multiplied by π£ with respect to π₯, which is the integral of negative one divided by two multiplied by two π₯ plus one all multiplied by three π to the power of two π₯ multiplied by one plus two π₯ with respect to π₯.

At first, this might not seem any simpler. However, we can notice in our integrand that our denominator and our numerator share a factor of one plus two π₯. So, we can simplify this to be negative three π₯π to the power of two π₯ divided by two multiplied by two π₯ plus one minus the integral of negative three π to the two π₯ over two with respect to π₯. We can now evaluate this integral by recalling that for contestants π and π, the integral of ππ to the ππ₯ with respect to π₯ is equal to ππ to the ππ₯ divided by π plus our constant of integration π.

So, evaluating the negative of this integral gives us three π to the power of two π₯ divided by four plus our constant of integration π. We can simplify this further by taking out a shared factor of three-quarters π to the power of two π₯ from our first and second term. This gives us three-quarters π to the power of two π₯ multiplied by negative two π₯ over two π₯ plus one plus one plus our constant of integration π.

We can then add the terms inside our parentheses by noticing that one is equal to two π₯ plus one over two π₯ plus one. Then, we see that the negative two π₯ and the two π₯ in our numerators cancel. So, weβve shown that our function π of π₯ is equal to three-quarters π to the power of two π₯ multiplied by one over two π₯ plus one plus our constant of integration π.

Finally, we recall that the question tells us that the point one, five π squared lies on the curve π¦ is equal to π of π₯. If the point one, five π squared lies on our curve π¦ is equal to π of π₯, then when π₯ is equal to one, π¦ must be equal to five π squared. So, we substitute π₯ is equal to one and π¦ is equal to five π squared into our equation. This gives us that five π squared is equal to three-quarters π to the power of two times one all multiplied by one over two times one plus one plus our constant of integration π.

The right-hand side of our equation simplifies to give us three-quarters π squared multiplied by a third plus π. We can then cancel the shared factors of three in our numerator and our denominator, giving us that five π squared is equal to π squared over four plus π. Finally, weβll subtract π squared over four from both sides of this equation to give us that π is equal to 19π squared over four.

Therefore, after a bit of rearranging, weβve shown that if the slope of the tangent to the curve π¦ is equal to π of π₯ at the point π₯, π¦ is given by three π₯π to the to two π₯ over two π₯ plus one squared. And our curve passes through the point one, five π squared. Then, our function π of π₯ is equal to three π to the power of two π₯ over four multiplied by two π₯ plus one plus 19π squared over four.