Question Number 171046 by Beginner last updated on 06/Jun/22
Answered by thfchristopher last updated on 07/Jun/22
![Let u=a+bx du=bdx, x=((u−a)/b) ∴∫(x^2 /((a+bx)^2 ))dx =(1/b^3 )∫(((u−a)^2 )/u^2 )du =(1/b^3 )∫((u^2 −2au+a^2 )/u^2 )du =(1/b^3 )(∫du−2a∫(1/u)du+a^2 ∫(1/u^2 )du) =(1/b^3 )(u−2aln u−(a^2 /u))+C =(1/b^3 )[(a+bx)+2aln (a+bx)−(a^2 /((a+bx)))]+C](https://www.tinkutara.com/question/Q171055.png)
$$\mathrm{Let}\:{u}={a}+{bx} \\ $$$${du}={bdx},\:{x}=\frac{{u}−{a}}{{b}} \\ $$$$\therefore\int\frac{{x}^{\mathrm{2}} }{\left({a}+{bx}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\int\frac{\left({u}−{a}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\int\frac{{u}^{\mathrm{2}} −\mathrm{2}{au}+{a}^{\mathrm{2}} }{{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\left(\int{du}−\mathrm{2}{a}\int\frac{\mathrm{1}}{{u}}{du}+{a}^{\mathrm{2}} \int\frac{\mathrm{1}}{{u}^{\mathrm{2}} }{du}\right) \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\left({u}−\mathrm{2}{a}\mathrm{ln}\:{u}−\frac{{a}^{\mathrm{2}} }{{u}}\right)+{C} \\ $$$$=\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\left[\left({a}+{bx}\right)+\mathrm{2}{a}\mathrm{ln}\:\left({a}+{bx}\right)−\frac{{a}^{\mathrm{2}} }{\left({a}+{bx}\right)}\right]+{C} \\ $$
Answered by floor(10²Eta[1]) last updated on 06/Jun/22
![I=∫((x/(a+bx)))^2 dx x=(a+bx)q+r=bqx+aq+r ⇒bq=1, aq+r=0 ⇒q=(1/b), r=−(a/b) I=∫((1/b)−(a/(b(a+bx))))^2 dx =∫((1/b^2 )−((2a)/(b^2 (a+bx)))+(a^2 /(b^2 (a+bx)^2 )))dx =(x/b^2 )−((2aln(a+bx))/b^3 )−(a^2 /b^3 )((1/(a+bx)))+C [easily done by letting u=a+bx]](https://www.tinkutara.com/question/Q171052.png)
$$\mathrm{I}=\int\left(\frac{\mathrm{x}}{\mathrm{a}+\mathrm{bx}}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{x}=\left(\mathrm{a}+\mathrm{bx}\right)\mathrm{q}+\mathrm{r}=\mathrm{bqx}+\mathrm{aq}+\mathrm{r} \\ $$$$\Rightarrow\mathrm{bq}=\mathrm{1},\:\mathrm{aq}+\mathrm{r}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{q}=\frac{\mathrm{1}}{\mathrm{b}},\:\mathrm{r}=−\frac{\mathrm{a}}{\mathrm{b}} \\ $$$$\mathrm{I}=\int\left(\frac{\mathrm{1}}{\mathrm{b}}−\frac{\mathrm{a}}{\mathrm{b}\left(\mathrm{a}+\mathrm{bx}\right)}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }−\frac{\mathrm{2a}}{\mathrm{b}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{bx}\right)}+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} \left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{2}} }\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{b}^{\mathrm{2}} }−\frac{\mathrm{2aln}\left(\mathrm{a}+\mathrm{bx}\right)}{\mathrm{b}^{\mathrm{3}} }−\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{3}} }\left(\frac{\mathrm{1}}{\mathrm{a}+\mathrm{bx}}\right)+\mathrm{C} \\ $$$$\left[\mathrm{easily}\:\mathrm{done}\:\mathrm{by}\:\mathrm{letting}\:\mathrm{u}=\mathrm{a}+\mathrm{bx}\right] \\ $$