Question Number 22221 by Sahib singh last updated on 13/Oct/17
$$\int\:\:\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} {x}^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by Sahib singh last updated on 16/Oct/17
$${thank}\:{you}\:{so}\:{much}. \\ $$
Answered by squidward last updated on 14/Oct/17
![=∫((a_0 +b_0 (a+x−a)^2 )/((a+x)^2 ))dx =∫((a_0 +b_0 (a+x)^2 −2b_0 (a+x)a+b_0 a^2 )/((a+x)^2 ))d(a+x) =∫[((a_0 +b_0 a^2 )/((a+x)^2 ))+b_0 −((2b_0 a)/(a+x))]d(a+x) =−((a_0 +b_0 a^2 )/(a+x))+b_0 (a+x)−2b_0 a ln∣a+x∣+C](Q22301.png)
$$=\int\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} \left({a}+{x}−{a}\right)^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$=\int\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} \left({a}+{x}\right)^{\mathrm{2}} −\mathrm{2}{b}_{\mathrm{0}} \left({a}+{x}\right){a}+{b}_{\mathrm{0}} {a}^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }{d}\left({a}+{x}\right) \\ $$$$=\int\left[\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} {a}^{\mathrm{2}} }{\left({a}+{x}\right)^{\mathrm{2}} }+{b}_{\mathrm{0}} −\frac{\mathrm{2}{b}_{\mathrm{0}} {a}}{{a}+{x}}\right]{d}\left({a}+{x}\right) \\ $$$$=−\frac{{a}_{\mathrm{0}} +{b}_{\mathrm{0}} {a}^{\mathrm{2}} }{{a}+{x}}+{b}_{\mathrm{0}} \left({a}+{x}\right)−\mathrm{2}{b}_{\mathrm{0}} {a}\:{ln}\mid{a}+{x}\mid+{C} \\ $$