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Question Number 97775 by liki last updated on 09/Jun/20

Commented by liki last updated on 09/Jun/20

$$..\mathrm{i}\mathrm{need}\mathrm{help}$$

Commented by Tony Lin last updated on 09/Jun/20

$$\int \frac{x^2}{x^2+bx+12}dx$$$$=\int \frac{x^2+bx+12}{x^2+bx+12}-\frac{b}{2}\int \frac{2x+b}{x^2+bx+12}dx$$$$-\frac{b}{2}\int \frac{\frac{24}{b}-b}{{(x+\frac{b}{2})}^2+{\left(\sqrt{12-\frac{b^2}{4}}\right)}^2}dx$$$$=x-\frac{b}{2}ln\mid x^2+bx+12\mid +\frac{b^2-24}{\sqrt{48-b^2}}{tan}^{-1}\frac{2x+b}{\sqrt{48-b^2}}+c$$

Answered by Ar Brandon last updated on 09/Jun/20

$$\mathcal{I}=\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}=\int \{1-\frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}\}$$$$=\int \{1-\frac{\mathrm{b}}{2}\cdot \frac{2\mathrm{x}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{bx}+12}+\frac{\frac{{\mathrm{b}}^2}{2}-12}{{\mathrm{x}}^2+\mathrm{bx}+12}\}\mathrm{dx}$$$$=\int \{1-\frac{\mathrm{b}}{2}\cdot \frac{2\mathrm{x}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{bx}+12}+2\cdot \frac{{\mathrm{b}}^2-24}{{(2\mathrm{x}+\mathrm{b})}^2+(48-{\mathrm{b}}^2)}\}\mathrm{dx}$$$$=\mathrm{x}-\frac{\mathrm{b}}{2}\cdot \ln \mid {\mathrm{x}}^2+\mathrm{bx}+12\mid +\frac{{\mathrm{b}}^2-24}{\sqrt{48-{\mathrm{b}}^2}}{\tan }^{-1}\left[\frac{2\mathrm{x}+\mathrm{b}}{\sqrt{48-{\mathrm{b}}^2}}\right]+\mathcal{C}$$

Answered by abdomathmax last updated on 09/Jun/20

$$\mathrm{I}=\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}\Rightarrow \mathrm{I}=\int \frac{{\mathrm{x}}^2+\mathrm{bx}+12-\mathrm{bx}-12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\mathrm{x}-\int \frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$${\mathrm{x}}^2+\mathrm{bx}+12=0\to \mathrm{\Delta }={\mathrm{b}}^2-48$$$$\mathrm{case}1{\mathrm{b}}^2-48>0\Rightarrow {\mathrm{x}}_1=\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^2-48}}{2}$$$${\mathrm{x}}_2=\frac{-\mathrm{b}-\sqrt{{\mathrm{b}}^2-48}}{2}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}=\frac{\mathrm{bx}+12}{(\mathrm{x}-{\mathrm{x}}_1)(\mathrm{x}-{\mathrm{x}}_2)}=\frac{\mathrm{a}}{\mathrm{x}-{\mathrm{x}}_1}+\frac{\mathrm{b}}{\mathrm{x}-{\mathbf{x}}_2}$$$$\mathbf{a}=\frac{{\mathbf{bx}}_1+12}{\sqrt{{\mathbf{b}}^2-48}}\mathbf{and}\mathbf{b}=-\frac{{\mathbf{bx}}_2+12}{\sqrt{{\mathbf{b}}^2-48}}$$$$\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{aln}\mid \mathrm{x}-{\mathrm{x}}_1\mid +\mathrm{bln}\mid \mathrm{x}-{\mathrm{x}}_2\mid +\mathrm{c}\Rightarrow$$$$\mathrm{I}=\mathrm{x}-\mathrm{aln}\mid \mathrm{x}-{\mathrm{x}}_1\mid -\mathrm{bln}\mid \mathrm{x}-{\mathrm{x}}_2\mid +\mathrm{C}$$$$\mathrm{case}2{\mathrm{b}}^2-48<0\Rightarrow \frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}$$$$=\frac{\mathrm{bx}+12}{{\mathrm{x}}^2+2\mathrm{x}\frac{\mathrm{b}}{2}+\frac{{\mathrm{b}}^2}{4}+12-\frac{{\mathrm{b}}^2}{4}}=\frac{\mathrm{bx}+12}{{(\mathrm{x}+\frac{\mathrm{b}}{2})}^2+\frac{48-{\mathrm{b}}^2}{4}}$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}+\frac{\mathrm{b}}{2}=\frac{\sqrt{48-{\mathrm{b}}^2}}{2}\mathrm{u}\Rightarrow$$$$\int \frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}=\frac{4}{\sqrt{48-{\mathrm{b}}^2}}\int \frac{\mathrm{b}(\frac{\sqrt{48-{\mathrm{b}}^2}}{2}\mathrm{u}-\frac{\mathrm{b}}{2})+12}{1+{\mathrm{u}}^2}\times \frac{\sqrt{48-{\mathrm{b}}^2}}{2}\mathrm{du}$$$$=2\int \frac{\sqrt{48-{\mathrm{b}}^2}\mathrm{u}-\frac{{\mathrm{b}}^2}{2}+12}{1+{\mathrm{u}}^2}\mathrm{du}$$$$=\sqrt{48-{\mathrm{b}}^2}\ln (1+{\mathrm{u}}^2)+(24-{\mathrm{b}}^2)\arctan \left(\mathrm{u}\right)+\mathrm{C}$$$$=\sqrt{48-{\mathrm{b}}^2}\ln (1+{\left(\frac{2\mathrm{x}+\mathrm{b}}{\sqrt{48-{\mathrm{b}}^2}}\right)}^2)$$$$+(24-{\mathrm{b}}^2)\arctan \left(\frac{2\mathrm{x}+\mathrm{b}}{\sqrt{48-{\mathrm{b}}^2}}\right)+\mathrm{C}\Rightarrow$$$$\mathrm{I}=\mathrm{x}-\sqrt{48-{\mathrm{b}}^2}\ln (1+\frac{{(2\mathrm{x}+\mathrm{b})}^2}{48-{\mathrm{b}}^2})$$$$-(24-{\mathrm{b}}^2)\arctan \left(\frac{2\mathrm{x}+\mathrm{b}}{\sqrt{48-{\mathrm{b}}^2}}\right)+\mathrm{C}$$

Answered by niroj last updated on 09/Jun/20

$$\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\int \frac{{\mathrm{x}}^2+\mathrm{bx}+12-\mathrm{bx}-12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\int \frac{{\mathrm{x}}^2+\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}-\int \frac{\mathrm{bx}+12}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\int \mathrm{dx}-\int \frac{\frac{\mathrm{b}}{2}(2\mathrm{x}+\mathrm{b})+12-\frac{{\mathrm{b}}^2}{2}}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\int \mathrm{dx}-\frac{\mathrm{b}}{2}\int \frac{(2\mathrm{x}+\mathrm{b})}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}-(12-\frac{{\mathrm{b}}^2}{2})\int \frac{1}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}$$$$=\int \mathrm{dx}-\frac{\mathrm{b}}{2}\int \frac{(2\mathrm{x}+\mathrm{b})}{{\mathrm{x}}^2+\mathrm{bx}+12}\mathrm{dx}-(12-\frac{{\mathrm{b}}^2}{2})\int \frac{1}{{\left(\mathrm{x}\right)}^2+2\mathrm{x}.\frac{\mathrm{b}}{2}+\frac{{\mathrm{b}}^2}{4}-\frac{{\mathrm{b}}^2}{4}+12}\mathrm{dx}$$$$=\int \mathrm{dx}-\frac{\mathrm{b}}{2}\int \frac{(2\mathrm{x}+\mathrm{b})\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{bx}+12}-(12-\frac{{\mathrm{b}}^2}{2})\int \frac{1}{{(\mathrm{x}+\frac{\mathrm{b}}{2})}^2-\left(\frac{{\mathrm{b}}^2-48}{4}\right)}\mathrm{dx}$$$$=\int \mathrm{dx}-\frac{\mathrm{b}}{2}\int \frac{(2\mathrm{x}+\mathrm{b})\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{bx}+12}-(12-\frac{{\mathrm{b}}^2}{2})\int \frac{1}{{(\mathrm{x}+\frac{\mathrm{b}}{2})}^2-{\left(\frac{\sqrt{{\mathrm{b}}^2-48}}{2}\right)}^2}\mathrm{dx}$$$$=\mathrm{x}-\frac{\mathrm{b}}{2}\log ({\mathrm{x}}^2+\mathrm{bx}+12)-(12-\frac{{\mathrm{b}}^2}{2}).\frac{1}{2.\frac{\sqrt{{\mathrm{b}}^2-48}}{2}}\log \frac{\mathrm{x}+\frac{\mathrm{b}}{2}-\frac{\sqrt{{\mathrm{b}}^2-48}}{2}}{\mathrm{x}+\frac{\mathrm{b}}{2}+\frac{\sqrt{{\mathrm{b}}^2-48}}{2}}+\mathrm{C}$$$$=\mathrm{x}-\frac{\mathrm{b}}{2}\log ({\mathrm{x}}^2+\mathrm{bx}+12)-(12-\frac{{\mathrm{b}}^2}{2})\frac{1}{\sqrt{{\mathrm{b}}^2-48}}\log \frac{2\mathrm{x}-\mathrm{b}-\sqrt{{\mathrm{b}}^2-48}}{2\mathrm{x}+\mathrm{b}+\sqrt{{\mathrm{b}}^2-48}}+\mathrm{C}//.$$$$$$

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