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Question Number 94161 by i jagooll last updated on 17/May/20

$$\int \frac{\mathrm{dx}}{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}$$

Commented by mathmax by abdo last updated on 17/May/20

$$I=\int \frac{dx}{p+\sqrt{qx+r}}weusethechangememt\sqrt{qx+r}=t\Rightarrow$$$$qx+r=t^2\Rightarrow qx=t^2-r\Rightarrow dx=\frac{2t}{q}dt\Rightarrow$$$$I=\frac{2}{q}\int \frac{tdt}{p+t}=\frac{2}{q}\int \frac{p+t-p}{p+t}dt=\frac{2}{q}t-\frac{2p}{q}\int \frac{dt}{t+p}$$$$=\frac{2}{q}t-\frac{2p}{q}ln\mid t+p\mid +C$$$$=\frac{2\sqrt{qx+r}}{q}-\frac{2p}{q}ln\mid p+\sqrt{qx+r}\mid +C(wesupposeq\ne 0)$$$$ifq=0\Rightarrow I=\frac{x}{p+\sqrt{r}}$$

Commented by hknkrc46 last updated on 17/May/20

$$\int \frac{\mathrm{dx}}{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}\{\begin{array}{l}\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}=\mathrm{t}\\ \mathrm{x}=\frac{{(\mathrm{t}-\mathrm{p})}^2-\mathrm{r}}{\mathrm{q}}\\ \mathrm{dx}=\frac{2(\mathrm{t}-\mathrm{p})}{\mathrm{q}}\mathrm{dt}\end{array}$$$$=\frac{2}{\mathrm{q}}\int \frac{\mathrm{t}-\mathrm{p}}{\mathrm{t}}\mathrm{dt}=\frac{2}{\mathrm{q}}\int (1-\frac{\mathrm{p}}{\mathrm{t}})\mathrm{dt}$$$$=\frac{2\mathrm{t}}{\mathrm{q}}-\frac{2\mathrm{p}}{\mathrm{q}}\ln \mid \mathrm{t}\mid +{\mathrm{C}}_1=-\ln \mid \mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}{\mid }^{\frac{2\mathrm{p}}{\mathrm{q}}}+\ln \mid \mathrm{C}\mid$$$$\{\ln \mid \mathrm{C}\mid ={\mathrm{C}}_1+\frac{2\mathrm{t}}{\mathrm{q}}\}$$$$=\ln \mid \frac{\mathrm{C}}{{(\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}})}^{\frac{2\mathrm{p}}{\mathrm{q}}}}\mid$$

Answered by john santu last updated on 17/May/20

$$\mathrm{set}\sqrt{\mathrm{qx}+\mathrm{r}}=\mathrm{pt}\Rightarrow \mathrm{qx}+\mathrm{r}={\mathrm{p}}^2{\mathrm{t}}^2$$$$\mathrm{dx}=\frac{{2\mathrm{p}}^2\mathrm{t}}{\mathrm{q}}\mathrm{dt}$$$$\mathrm{I}=\int \frac{{2\mathrm{p}}^2\mathrm{t}\mathrm{dt}}{\mathrm{q}(\mathrm{p}+\mathrm{pt})}=\frac{2\mathrm{p}}{\mathrm{q}}\int \frac{\mathrm{t}}{1+\mathrm{t}}\mathrm{dt}$$$$\mathrm{I}=\frac{2\mathrm{p}}{\mathrm{q}}\int \frac{\mathrm{t}+1-1}{\mathrm{t}+1}\mathrm{dt}=\frac{2\mathrm{p}}{\mathrm{q}}(\mathrm{t}-\ln \mid \mathrm{t}+1\mid +\mathrm{c}$$$$\mathrm{I}=\frac{2\mathrm{p}}{\mathrm{q}}\mathrm{t}-\frac{2\mathrm{p}}{\mathrm{q}}\ln \mid \mathrm{t}+1\mid +\mathrm{c}$$$$\mathrm{I}=\frac{2\mathrm{p}\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{pq}}-\frac{2\mathrm{p}}{\mathrm{q}}\ln \mid \frac{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{p}}\mid +\mathrm{c}$$$$\mathrm{I}=\frac{2\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{q}}-\frac{2\mathrm{p}}{\mathrm{q}}\ln \mid \frac{\mathrm{p}+\sqrt{\mathrm{qx}+\mathrm{r}}}{\mathrm{p}}\mid +\mathrm{c}$$

Commented by i jagooll last updated on 17/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 04/Jun/20

$$\mathrm{thank}\mathrm{you}$$

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