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Use integration by parts to derive the following formula from the table of integrals.$$\int x^{n} \cdot \ln |x| d x=x^{n+1}\left[\frac{\ln |x|}{n+1}-\frac{1}{(n+1)^{2}}\right]+C, \quad n \neq-1$$

$$\int x^{n} . \ln |x| d x=x^{n+1}\left[\frac{\ln |x|}{n+1}-\frac{1}{(n+1)^{2}}\right]+C, \mathrm{n} \neq-1$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 8

Further Techniques and Applications of Integration

Section 1

Integration by Parts

Integrals

Integration

Integration Techniques

Applications of Integration

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problem. We're being asked to take the integral of X to the end times the natural log of X, using integration by parts to derive the formula that's in the table of integral Sze. So using for the integration by parts, we're gonna let you be the natural log of X. That means the derivative is going to be one over x dx and we'll let Devi b X to the n DX, which means V is going to be won over and plus one times X to the n plus one so are integral of X to the end times Natural law giver of civility of X d. X is equal to UV so one over and plus one times X to the n plus one times the natural log of X minus the integral of VD You so that's going to be one over and plus one, uh, noticed that the X to the n plus one and the ex will cancel one of them out will have next to the end left over DX, so that first part will combine to X to the n plus one over M plus one times the natural log of absolute value of X and the integral itself. This is we're gonna be bringing the end down after we increased the exponents by one is this is just a number to the n plus one comes down, we'll get one over and plus one squared X to the power of and plus one plus c. And the formula in the back has X to the n plus one factor. Don't. So we're left with natural log of absolutely of X over and plus one minus one over and plus one squared plus c That's the formula in the back.

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