# Question Video: Finding the Integration of a Function by Distributing the Division and Applying the Power Rule Mathematics

Determine β« ((7π₯Β² β 3)/6π₯) dπ₯.

02:24

### Video Transcript

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

In this question, weβre asked to evaluate the integral of a rational function. And this is very difficult to integrate in its current form. So instead, letβs simplify this expression. To do this, weβll 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. Letβs 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 ππ₯ to the πth 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 πΆ. 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.

Letβs 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 weβre 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 πΆ. 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 πΆ. 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 πΆ.