How would I use the Comparison Test in calculus to determine that the integral dx/ln(x) diverges over 2 to infinity?
Also, I need to solve the integral (cos(4x) +3) / (3x^2 + 1) over 1 to infinity using the comparison test as well.
If you could help on one or both, that would be lovely. Thanks!
Also, I need to solve the integral (cos(4x) +3) / (3x^2 + 1) over 1 to infinity using the comparison test as well.
If you could help on one or both, that would be lovely. Thanks!
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When x is at least 2, ln(x) is less than x, so 1/ln(x) is greater than 1/x. We know that the integral of dx/x from 2 to infinity diverges, so by the comparison test, so does the integral of dx/ln(x) from 2 to infinity.
For your second question, we know that cos(4x) is always between -1 and 1, so cos(4x)+3 is between 2 and 4, and is hence less than 5. Moreover, when x is at least 1, 3x^2+1 is greater than x^2. So, (cos(4x)+3)/(3x^2+1) is less than 5/x^2. We know that the integral of dx/x^2 (and hence 5dx/x^2) from 1 to infinity converges, so by the comparison test, so does the integral of (cos(4x)+3)/(3x^2+1)dx from 1 to infinity.
Hope that helps! :)
For your second question, we know that cos(4x) is always between -1 and 1, so cos(4x)+3 is between 2 and 4, and is hence less than 5. Moreover, when x is at least 1, 3x^2+1 is greater than x^2. So, (cos(4x)+3)/(3x^2+1) is less than 5/x^2. We know that the integral of dx/x^2 (and hence 5dx/x^2) from 1 to infinity converges, so by the comparison test, so does the integral of (cos(4x)+3)/(3x^2+1)dx from 1 to infinity.
Hope that helps! :)