Question #233061

True or False

To evaluate the integral xln(x)dx\displaystyle{\int \sqrt{x}\ln{(x)}dx}  by integration by parts,  it is wise to choose u=xu=\sqrt{x} and dv=ln(x)dxdv=\ln{(x)}dx



1
Expert's answer
2021-09-06T15:00:46-0400

False


 it is not wise to choose u=xu=\sqrt{x} and dv=ln(x)dxdv=\ln{(x)}dx .

Since the anti derivative of x\sqrt{x} , is known but the anti derivative of lnxlnx is not. So u=ln(x)u=ln(x) and dv=xdxdv=\sqrt{x}dx

du=1xdu=\frac{1}{x}


v=2x323v=\frac{2x^{\frac{3}{2}}}{3}


This because the derivative of ln(x)ln(x) which is 1x\frac{1}{x} , will interact nicely with polynomial terms, whereas the integral is some weird derivative on the order of xlnxxlnx which looks terrible






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