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Question Number 117666 by danielasebhofoh last updated on 13/Oct/20

Commented by bemath last updated on 13/Oct/20

$$\mathrm{set}\sqrt{\mathrm{ax}+\mathrm{b}}=\mathrm{t}\Rightarrow \mathrm{ax}+\mathrm{b}={\mathrm{t}}^2$$$$\mathrm{x}=\frac{{\mathrm{t}}^2-\mathrm{b}}{\mathrm{a}}\Rightarrow \mathrm{dx}=\frac{2\mathrm{t}\mathrm{dt}}{\mathrm{a}}$$$$\int \frac{2\mathrm{t}\mathrm{dt}}{\mathrm{a}(\frac{{\mathrm{t}}^2-\mathrm{b}}{\mathrm{a}}+\mathrm{c}).\mathrm{t}}=\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2-(\mathrm{b}-\mathrm{ac})}$$$$\int \frac{2\mathrm{dt}}{(\mathrm{t}-\sqrt{\mathrm{b}-\mathrm{ac}})(\mathrm{t}+\sqrt{\mathrm{b}-\mathrm{ac}})}$$$$\mathrm{now}\mathrm{it}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

Commented by MJS_new last updated on 13/Oct/20

$$\mathrm{typo}:\int \frac{2dt}{t^2-(b-ac)}$$

Commented by bemath last updated on 13/Oct/20

$$\mathrm{haha}..\mathrm{yes}\mathrm{sir}\mathrm{thank}\mathrm{you}$$

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