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

Determine ∫ ((7π‘₯Β² βˆ’ 3)/6π‘₯) dπ‘₯.

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Video Transcript

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

In this question, we’re asked to evaluate the integral of a rational function. And this is very difficult to integrate in its current form. So instead, let’s simplify this expression. To do this, we’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. Let’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‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Žπ‘₯ to the 𝑛th power with respect to π‘₯ is π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus a constant of integration 𝐢. 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.

Let’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 we’re just integrating a constant multiple of the reciprocal function. And we can do this by recalling for any real constant π‘Ž, the integral of π‘Ž over π‘₯ with respect to π‘₯ is equal to π‘Ž times the natural logarithm of the absolute value of π‘₯ plus a constant of integration 𝐢. And in our case, the value of π‘Ž 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 𝐢. 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 𝐢.

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