Originally posted by lecture

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∫x³ ln(2+x²)dx:

-> ∫x² ln(2+x²) x dx

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(Substitution)
w=2+x²
dw=2x dx
dw/2= x dx
x²=w-2
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-> ∫(1/2)(w-2) ln w dw
-> ∫[(w/2)-1] ln w dw
-> ∫(w/2) ln w - ln w dw
-> (1/2)∫w ln w dw - ∫ln w dw

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(Parts)
u= ln w, dv= w dw
du= dw/w, v= w²/2

∫w ln w = uv-∫v du = (1/2) w² ln w - (1/2)∫w = (1/2) w² ln w - w²/4 + C
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∫x³ ln(2+x²)dx = (1/4) w² ln w - w²/8 - w(ln w - 1) +C

== [(2+x²)² ln (2+x²)/4] - [(2+x²)²/8] - (2+x²)[ln (2+x²) - 1] +C

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