How do you find the integral of $(x^3 + 1) lnxdx$ ?

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Answer 1 · 2018-04-30T00:56:29

$(x^4/4+x)(In x)-x^4/16+x+C$

Let u= $In x$
And dv= $x^3+1$ SO v= $(x^4/4+x)$

Integration by parts is given by = $vu-intv du$

$int (x^3+1) In x dx$ = $(x^4/4+x)(In x)-int (x)(x^3/4+1)times1/x dx$

= $(x^4/4+x)(In x)-int (x^3/4+1) dx$

= $(x^4/4+x)(In x)-x^4/16+x+C$

How do you find the integral of (x^3 + 1) lnxdx(x^3 + 1) lnxdx?