Question

# Integrate the function. ∫x(logx)2dx

Solution

## Let I=∫x(logx)2dx On taking (logx)2 as first function and x as second function and integrating by parts, we get I=(logx)2∫xdx−∫[ddx(logx)2∫xdx]dx=(logx)2.x22−∫[2logxx.x22]dx=x22(logx)2−∫xlogxdx Again integrating by parts, we get I=x22(logx)2−[logx∫xdx−∫{(ddxlogx)∫xdx}dx]+C=x22(logx)2−[x22logx−∫1x.x22dx]+C=x22(logx)2−x22logx+12∫xdx+C=x22(logx)2−x22logx+x24+C Mathematics

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