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Question Number 171046 by Beginner last updated on 06/Jun/22

Answered by thfchristopher last updated on 07/Jun/22

$$\mathrm{Let}u=a+bx$$$$du=bdx,x=\frac{u-a}{b}$$$$\therefore \int \frac{x^2}{{(a+bx)}^2}dx$$$$=\frac{1}{b^3}\int \frac{{(u-a)}^2}{u^2}du$$$$=\frac{1}{b^3}\int \frac{u^2-2au+a^2}{u^2}du$$$$=\frac{1}{b^3}(\int du-2a\int \frac{1}{u}du+a^2\int \frac{1}{u^2}du)$$$$=\frac{1}{b^3}(u-2a\ln u-\frac{a^2}{u})+C$$$$=\frac{1}{b^3}[(a+bx)+2a\ln (a+bx)-\frac{a^2}{(a+bx)}]+C$$

Answered by floor(10²Eta[1]) last updated on 06/Jun/22

$$\mathrm{I}=\int {\left(\frac{\mathrm{x}}{\mathrm{a}+\mathrm{bx}}\right)}^2\mathrm{dx}$$$$\mathrm{x}=(\mathrm{a}+\mathrm{bx})\mathrm{q}+\mathrm{r}=\mathrm{bqx}+\mathrm{aq}+\mathrm{r}$$$$\Rightarrow \mathrm{bq}=1,\mathrm{aq}+\mathrm{r}=0$$$$\Rightarrow \mathrm{q}=\frac{1}{\mathrm{b}},\mathrm{r}=-\frac{\mathrm{a}}{\mathrm{b}}$$$$\mathrm{I}=\int {(\frac{1}{\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{b}(\mathrm{a}+\mathrm{bx})})}^2\mathrm{dx}$$$$=\int (\frac{1}{{\mathrm{b}}^2}-\frac{2\mathrm{a}}{{\mathrm{b}}^2(\mathrm{a}+\mathrm{bx})}+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2{(\mathrm{a}+\mathrm{bx})}^2})\mathrm{dx}$$$$=\frac{\mathrm{x}}{{\mathrm{b}}^2}-\frac{2\mathrm{aln}(\mathrm{a}+\mathrm{bx})}{{\mathrm{b}}^3}-\frac{{\mathrm{a}}^2}{{\mathrm{b}}^3}\left(\frac{1}{\mathrm{a}+\mathrm{bx}}\right)+\mathrm{C}$$$$[\mathrm{easily}\mathrm{done}\mathrm{by}\mathrm{letting}\mathrm{u}=\mathrm{a}+\mathrm{bx}]$$

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