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Question Number 107756 by bemath last updated on 12/Aug/20

$$\frac{\mathcal{B}e\mathcal{M}ath}{\coprod }$$$$\int x^2\ln (x^2+3)dx$$

Answered by hgrocks last updated on 12/Aug/20

$$$$$$\mathrm{I}=\ln ({\mathrm{x}}^2+3)\frac{{\mathrm{x}}^3}{3}-\frac{2}{3}\int \frac{{\mathrm{x}}^4}{{\mathrm{x}}^2+3}\mathrm{dx}\to \mathrm{J}$$$$\mathrm{J}=\int \frac{{\mathrm{x}}^4-9+9}{{\mathrm{x}}^2+3}\mathrm{dx}$$$$=\int ({\mathrm{x}}^2-3)\mathrm{dx}+9\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+3}$$$$=\frac{{\mathrm{x}}^3}{3}-3\mathrm{x}+\frac{9}{\sqrt{3}}{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)$$$$\mathrm{So}\mathrm{I}=\frac{{\mathrm{x}}^3}{9}(3\ln ({\mathrm{x}}^2+3)-2)+2(\mathrm{x}-\sqrt{3}{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right))$$

Answered by bobhans last updated on 12/Aug/20

$$\frac{⊺\mathcal{B}\mathrm{obhans}⊺}{\mathrm{\Pi }}$$$$\mathrm{set}\ln ({\mathrm{x}}^2+3)=\mathrm{u}\Rightarrow \frac{2\mathrm{x}}{{\mathrm{x}}^2+3}\mathrm{dx}=\mathrm{du}$$$$\mathrm{v}=\frac{1}{3}{\mathrm{x}}^3$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{1}{3}\int \frac{{2\mathrm{x}}^4}{({\mathrm{x}}^2+3)}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{2}{3}\int \frac{{\mathrm{x}}^4}{{\mathrm{x}}^2+3}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{2}{3}\int \frac{{({\mathrm{x}}^2+3)}^2-{6\mathrm{x}}^2-9}{{\mathrm{x}}^2+3}\mathrm{dx}$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{2}{3}\int (({\mathrm{x}}^2+3)-\frac{6({\mathrm{x}}^2+3)+9}{{\mathrm{x}}^2+3})\mathrm{dx}$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{2}{3}[\frac{1}{3}{\mathrm{x}}^3+3\mathrm{x}-6\mathrm{x}+\int \frac{9}{{\mathrm{x}}^2+3}\mathrm{dx}]$$$$\mathrm{I}=\frac{1}{3}{\mathrm{x}}^3\ln ({\mathrm{x}}^2+3)-\frac{2}{9}{\mathrm{x}}^3+2\mathrm{x}-2\sqrt{3}{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{3}}\right)+\mathrm{C}$$$$$$

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