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dx-p-qx-r-

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Question Number 94161 by i jagooll last updated on 17/May/20
∫ (dx/(p+(√(qx+r))))
$$\int\:\frac{\mathrm{dx}}{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}\: \\ $$
Commented by mathmax by abdo last updated on 17/May/20
I =∫ (dx/(p+(√(qx+r)))) we use the changememt (√(qx+r))=t ⇒ qx+r =t^2 ⇒qx =t^2 −r ⇒dx =((2t)/q) dt ⇒ I =(2/q)∫ ((tdt)/(p+t)) =(2/q)∫ ((p+t−p)/(p+t))dt =(2/q)t −((2p)/q) ∫ (dt/(t+p)) =(2/q)t −((2p)/q)ln∣t+p∣ +C =((2(√(qx+r)))/q) −((2p)/q)ln∣p+(√(qx+r))∣ +C (we suppose q≠0) if q=0 ⇒I =(x/(p+(√r)))
$${I}\:=\int\:\:\frac{{dx}}{{p}+\sqrt{{qx}+{r}}}\:\:{we}\:{use}\:{the}\:{changememt}\:\sqrt{{qx}+{r}}={t}\:\Rightarrow \\ $$$${qx}+{r}\:={t}^{\mathrm{2}} \:\Rightarrow{qx}\:={t}^{\mathrm{2}} −{r}\:\Rightarrow{dx}\:=\frac{\mathrm{2}{t}}{{q}}\:{dt}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{2}}{{q}}\int\:\:\frac{{tdt}}{{p}+{t}}\:=\frac{\mathrm{2}}{{q}}\int\:\frac{{p}+{t}−{p}}{{p}+{t}}{dt}\:=\frac{\mathrm{2}}{{q}}{t}\:−\frac{\mathrm{2}{p}}{{q}}\:\int\:\frac{{dt}}{{t}+{p}} \\ $$$$=\frac{\mathrm{2}}{{q}}{t}\:−\frac{\mathrm{2}{p}}{{q}}{ln}\mid{t}+{p}\mid\:+{C} \\ $$$$=\frac{\mathrm{2}\sqrt{{qx}+{r}}}{{q}}\:−\frac{\mathrm{2}{p}}{{q}}{ln}\mid{p}+\sqrt{{qx}+{r}}\mid\:+{C}\:\:\:\left({we}\:{suppose}\:{q}\neq\mathrm{0}\right) \\ $$$${if}\:{q}=\mathrm{0}\:\Rightarrow{I}\:=\frac{{x}}{{p}+\sqrt{{r}}} \\ $$
Commented by hknkrc46 last updated on 17/May/20
∫(dx/(p+(√(qx+r)))) { ((p+(√(qx+r))=t)),((x=(((t−p)^2 −r)/q))),((dx=((2(t−p))/q)dt)) :} =(2/q)∫ ((t−p)/t)dt=(2/q)∫(1−(p/t))dt =((2t)/q)−((2p)/q) ln ∣t∣+C_1 =−ln ∣p+(√(qx+r))∣^((2p)/q) + ln ∣C∣ {ln ∣C∣=C_1 +((2t)/q)} =ln ∣(C/((p+(√(qx+r)))^((2p)/q) ))∣
$$\int\frac{\mathrm{dx}}{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}\:\begin{cases}{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}=\mathrm{t}}\\{\mathrm{x}=\frac{\left(\mathrm{t}−\mathrm{p}\right)^{\mathrm{2}} −\mathrm{r}}{\mathrm{q}}}\\{\mathrm{dx}=\frac{\mathrm{2}\left(\mathrm{t}−\mathrm{p}\right)}{\mathrm{q}}\mathrm{dt}}\end{cases} \\ $$$$=\frac{\mathrm{2}}{\mathrm{q}}\int\:\frac{\mathrm{t}−\mathrm{p}}{\mathrm{t}}\mathrm{dt}=\frac{\mathrm{2}}{\mathrm{q}}\int\left(\mathrm{1}−\frac{\mathrm{p}}{\mathrm{t}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{2t}}{\mathrm{q}}−\frac{\mathrm{2p}}{\mathrm{q}}\:\mathrm{ln}\:\mid\mathrm{t}\mid+\mathrm{C}_{\mathrm{1}} =−\mathrm{ln}\:\mid\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}\mid^{\frac{\mathrm{2p}}{\mathrm{q}}} +\:\mathrm{ln}\:\mid\mathrm{C}\mid \\ $$$$\left\{\mathrm{ln}\:\mid\mathrm{C}\mid=\mathrm{C}_{\mathrm{1}} +\frac{\mathrm{2t}}{\mathrm{q}}\right\} \\ $$$$=\mathrm{ln}\:\mid\frac{\mathrm{C}}{\left(\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}\right)^{\frac{\mathrm{2p}}{\mathrm{q}}} }\mid \\ $$
Answered by john santu last updated on 17/May/20
set (√(qx+r)) = pt ⇒qx+r = p^2 t^2 dx = ((2p^2 t)/q) dt I= ∫ ((2p^2 t dt)/(q(p+pt))) = ((2p)/q)∫ (t/(1+t)) dt I = ((2p)/q) ∫ ((t+1−1)/(t+1)) dt = ((2p)/q) (t−ln∣t+1∣ +c I = ((2p)/q) t − ((2p)/q) ln∣t+1∣ + c I = ((2p (√(qx+r)))/(pq)) −((2p)/q) ln ∣ ((p+(√(qx+r)))/p)∣ + c I = ((2 (√(qx+r)))/q)− ((2p)/q) ln ∣((p+(√(qx+r)))/p)∣ + c
$$\mathrm{set}\:\sqrt{\mathrm{qx}+\mathrm{r}}\:=\:\mathrm{pt}\:\Rightarrow\mathrm{qx}+\mathrm{r}\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{dx}\:=\:\frac{\mathrm{2p}^{\mathrm{2}} \mathrm{t}}{\mathrm{q}}\:\mathrm{dt} \\ $$$$\mathrm{I}=\:\int\:\frac{\mathrm{2p}^{\mathrm{2}} \:\mathrm{t}\:\mathrm{dt}}{\mathrm{q}\left(\mathrm{p}+\mathrm{pt}\right)}\:=\:\frac{\mathrm{2p}}{\mathrm{q}}\int\:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}}\:\mathrm{dt}\: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{2p}}{\mathrm{q}}\:\int\:\frac{\mathrm{t}+\mathrm{1}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\:\mathrm{dt}\:=\:\frac{\mathrm{2p}}{\mathrm{q}}\:\left(\mathrm{t}−\mathrm{ln}\mid\mathrm{t}+\mathrm{1}\mid\:+\mathrm{c}\right. \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{2p}}{\mathrm{q}}\:\mathrm{t}\:−\:\frac{\mathrm{2p}}{\mathrm{q}}\:\mathrm{ln}\mid\mathrm{t}+\mathrm{1}\mid\:+\:\mathrm{c}\: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{2p}\:\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{pq}}\:−\frac{\mathrm{2p}}{\mathrm{q}}\:\mathrm{ln}\:\mid\:\frac{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{p}}\mid\:+\:\mathrm{c}\: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{2}\:\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{q}}−\:\frac{\mathrm{2p}}{\mathrm{q}}\:\mathrm{ln}\:\mid\frac{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{p}}\mid\:+\:\mathrm{c}\: \\ $$
Commented by i jagooll last updated on 17/May/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by peter frank last updated on 04/Jun/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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