### Video Transcript

Determine the integral of seven π₯
squared minus three all divided by six π₯ with respect to π₯.

In this question, weβre asked to
evaluate the integral of a rational function. And this is very difficult to
integrate in its current form. So instead, letβs simplify this
expression. To do this, weβll divide each term
in our numerator by the denominator separately. This gives us the integral of seven
π₯ squared over six π₯ minus three divided by six π₯ with respect to π₯. Now we can simplify our
integrand. In our first term, we can cancel
the shared factor of π₯. And in our second term, we can
cancel the shared factor of three in the numerator and denominator. This gives us the integral of
seven-sixths π₯ minus one over two π₯ with respect to π₯.

And now we can integrate each of
these separately. Letβs start with our first
term. We can see this is a polynomial
function. And we can integrate polynomial
functions by using the power rule for integration, which tells us for any real
constants π and π, where π is not equal to negative one, the integral of ππ₯ to
the πth power with respect to π₯ is π times π₯ to the power of π plus one divided
by π plus one plus a constant of integration πΆ. We add one to our exponent of π₯
and then divide by this new exponent. And in this function, we can write
the exponent of π₯ as one. So we add one to this exponent of
π₯ to get an exponent of two and divide by this new exponent of two. This gives us seven over six
multiplied by π₯ squared over two.

Letβs now move on to integrating
our second term. To do this, it might be easier to
rewrite this term as negative one-half multiplied by one over π₯. Then we can see weβre just
integrating a constant multiple of the reciprocal function. And we can do this by recalling for
any real constant π, the integral of π over π₯ with respect to π₯ is equal to π
times the natural logarithm of the absolute value of π₯ plus a constant of
integration πΆ. And in our case, the value of π is
negative one-half. So the integral of this term is
negative one-half times the natural logarithm of the absolute value of π₯. And remember, we need to add our
constant of integration πΆ. And we can simplify the first sum
of this expression to get our final answer. Seven over six times π₯ squared
over two is equal to seven π₯ squared over 12.

Therefore, we were able to show the
integral of seven π₯ squared minus three all divided by six π₯ with respect to π₯ is
equal to seven π₯ squared over 12 minus one-half times the natural logarithm of the
absolute value of π₯ plus the constant of integration πΆ.