Question

In Exercises 55 and \(56,\) use a computer algebra system to find the integral for \(n=0,1,2,\) and \(3 .\) Use the result to obtain a general rule for finding the integral for any positive integer \(n\), and test your results for \(n=4\). $$ \int x^{n} \ln x d x $$

Short answer

The integral of \(x^{n} \ln x dx\) is \[x^{n+1} \ln x - \frac{n+1}{n+2} x^{n+1} + C,\] for any positive integer \(n.\) For \(n=4,\) the integral is \[x^{5} \ln x - \frac{5}{6} x^{5} + C.\]

Step-by-step solution

Calculate the integral for \(n=0\)

Let's compute the integral when \(n=0:\) \[\int x^{0} \ln x dx.\] Since \(x^{0}=1,\) the integral becomes \[\int \ln x dx.\] Using integration by parts, where the standard format is \[\int u dv = u v - \int v du,\] let \(\ln x = u\) and \(dx = dv.\) Deriving these, we get \(du = \frac{1}{x} dx\) and \(v = x.\) So the integral becomes \[x \ln x - \int x \frac{1}{x} dx,\] which simplifies to \[x \ln x - x + C.\]

Calculate the integral for \(n=1\)

Now, let's compute the integral when \(n = 1:\) \[\int x^{1} \ln x dx = \int x \ln x dx.\] Apply integration by parts, we let \(x=u\) and \(\ln x dv.\) Getting the derivatives gives \(du = dx\) and \(v = x \ln x - x\). Thus, the integral is \[x^{2} \ln x - \frac{1}{2} x^{2} + C.\]

Calculate the integral for \(n=2\)

For \(n = 2:\) we have \[\int x^{2} \ln x dx.\] Applying integration by parts, we let \(x^{2}=u\) and \(\ln x dv.\) Deriving these gives \(du = 2x dx\) and \(v = x \ln x - x\). The integral becomes \[x^{3} \ln x - \frac{3}{4} x^{3} + C.\]

Calculate the integral for \(n=3\)

For \(n = 3:\) we will evaluate \[\int x^{3} \ln x dx.\] Using integration by parts and applying same logic as previous steps, we get \[x^{4} \ln x - \frac{4}{5} x^{4} + C.\]

Establish the generalized rule

Looking at the pattern from steps 1 to 4, we can discern a general rule for the integral of \(x^{n}\ln x dx.\) As a general rule, we find \[x^{n+1} \ln x - \frac{n+1}{n+2} x^{n+1} + C,\] where \(C\) is the constant of integration.

Test the generalized rule for \(n=4\)

Now, we test this rule for \(n=4.\) According to the rule, the integral will be \[x^{5} \ln x - \frac{5}{6} x^{5} + C.\]