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Numerade Educator

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Problem 35 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int \frac{5x^4 + 7x^2 + x + 2}{x (x^2 + 1)^2}\ dx $

Answer

$$2 \ln |x|+\frac{3}{2} \ln \left(x^{2}+1\right)+\frac{1}{2} \tan ^{-1} x+\frac{x}{2\left(x^{2}+1\right)}+C$$

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Video Transcript

So for this problem problem 35. What we're going to be doing is evaluating an integral, um through decomposition, partial fractions. So what we're given is five x to the fourth plus seven X squared plus X plus two over X times X squared plus one squared DX. So, through the partial composition of fractions, how we write this, uh, we're just going to look at the fraction. Now we're gonna worry about the integral later. What we'll have is a over X plus BX plus C over X squared, plus one and then we'll have plus D X plus e over X squared, plus one squared. Then what we do is we multiply, um, everything by the common denominator. So what we end up getting as a result is, um five x to the fourth plus seven x squared plus X plus two equals a Times X squared plus one squared plus x times b x plus C times x squared plus one plus, uh, x time's d X plus e. Then we see that we can categorize we can group everything together and now we get it into the form and that this same thing is equal to a plus b x to the fourth plus c x to the third plus to a plus B plus d x squared plus C plus e x plus. Okay, then we let the left side coefficient equal the right side coefficients. So we have the A plus b equals five c zero a to A plus B plus D equal seven c plus e equals one and two equals a well, based on that, since we know that we have that A is going to equal to C equals zero e is going to equal one because if this is just zero, then he obviously equals one. That's also gonna mean that b equals three because we put it to here. Subtracting at V equals three. Then that means that to a B Sorry to a is four b is three. So at seven, so D is just going to equal zero as well. So with these values in mind, we now go back to our original, um, set up, which is right here and now. What we're going to do is we're gonna have to over X plus three x over X squared plus one plus one over X squared plus one squared. Then we're going to take the derivative of that with respect to X. So we have two times the inner role of one over X DX, plus three times the inter role of X over X squared plus one plus the integral of one over X squared plus one squared DX. Our first in the room is going to give us to natural log of X plus C one. Our second integral is going to give us three halves natural log of X squared plus one plus C two and then our third one is going to give us, um, one half times the inverse tangent of X plus X over two times X squared plus one plus C three. So we combine all these together and our final answer is going to be to natural log X plus three halves natural log X squared plus one plus one half inverse tangent of acts plus x over two times X squared plus one plus C final final answer