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Question Number 185951 by manxsol last updated on 30/Jan/23

∫x^3 ln(x+4)dx

$$\int{x}^{\mathrm{3}} {ln}\left({x}+\mathrm{4}\right){dx} \\ $$

Answered by SEKRET last updated on 30/Jan/23

∫ x^3 ∙ln(x+4) dx= =(x^4 /4)∙ln(x+4) − ∫ (x^4 /4)∙(1/(x+4))dx = (x^4 /4)∙ln(x+4)−(1/4)∙∫(x^3 −4x^2 +16x+((256)/(x+4))−64)dx =(x^4 /4)∙ln(x−4)−(1/4)∙((x^4 /4)−((4x^3 )/3)+8x^2 +256∙ln(x+4)−64x)+C

$$\:\int\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} \centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)\:\boldsymbol{\mathrm{dx}}= \\ $$$$\:\:=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)\:−\:\int\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{4}}\boldsymbol{\mathrm{dx}} \\ $$$$=\:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{4}}\centerdot\int\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{16}\boldsymbol{\mathrm{x}}+\frac{\mathrm{256}}{\boldsymbol{\mathrm{x}}+\mathrm{4}}−\mathrm{64}\right)\boldsymbol{\mathrm{dx}} \\ $$$$=\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}−\mathrm{4}\right)−\frac{\mathrm{1}}{\mathrm{4}}\centerdot\left(\frac{\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{256}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)−\mathrm{64}\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{C}} \\ $$

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