Video: APCALC03AB-P1A-Q25-520108701281

Find ∫((2π‘₯⁴ βˆ’ 3π‘₯Β² + 5π‘₯)/π‘₯Β³) dπ‘₯.

02:54

Video Transcript

Find the integral of two π‘₯ to the fourth power minus three π‘₯ squared plus five π‘₯ all over π‘₯ cubed with respect to π‘₯.

Now, at first glance, this integral may look a little complex because it is a quotient. However, it’s just a quotient of polynomial functions in π‘₯. And the denominator of this quotient is simply π‘₯ cubed. We can split the integrand into the sum of three terms. It’s the integral of two π‘₯ to the fourth power over π‘₯ cubed minus three π‘₯ squared over π‘₯ cubed plus five π‘₯ over π‘₯ cubed with respect to π‘₯. And we can simplify each term by recalling our laws of indices.

We recall that in order to divide powers of the same base, in this case π‘₯, we subtract the powers. So, for our first term, we have two multiplied by π‘₯ to the power of four minus three. That’s just π‘₯ to the power of one. For our second term, we have negative three multiplied by π‘₯ to the power of two minus three. That’s negative one. And then, we have plus five multiplied by π‘₯ to the power of one minus three, which is negative two. And we’re integrating with respect to π‘₯.

We can now go ahead and perform this integral term-by-term. We need to recall two general rules of integration. Firstly, if we’re integrating π‘₯ to the power of some real number 𝑛, where 𝑛 is not equal to negative one, this is equal to π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus a constant. We increase the power by one and then divide by the new power. This will enable us to integrate both the first and the third terms.

But in the second term, we see that the power is equal to negative one. So, we need an alternative rule here. We recall that when 𝑛 is equal to negative one, the integral of π‘₯ to the power of negative one with respect to π‘₯ is equal to the natural logarithm of π‘₯ plus a constant.

So, let’s perform our integral. The integral of two π‘₯ with respect to π‘₯ is equal to two π‘₯ squared over two. The integral of negative three multiplied by π‘₯ to the power of negative one is equal to negative three multiplied by the natural logarithm of π‘₯. And the integral of positive five multiplied by π‘₯ to the power of negative two is positive five multiplied by π‘₯ to the power of negative one over negative one. And we’ll include one constant of integration 𝑐.

We can cancel a factor of two in our first term and rewrite our third term as a reciprocal to give our final answer. The integral of two π‘₯ to the fourth power minus three π‘₯ squared plus five π‘₯ all over π‘₯ cubed with respect to π‘₯ is equal to π‘₯ squared minus three times the natural logarithm of π‘₯ minus five over π‘₯ plus 𝑐.

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