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Question Number 97082 by bobhans last updated on 06/Jun/20

Answered by john santu last updated on 06/Jun/20

Commented by bobhans last updated on 06/Jun/20

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

Answered by mathmax by abdo last updated on 06/Jun/20

$$\mathrm{I}=\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{a}+\mathrm{bx}}}\mathrm{changement}\sqrt{\mathrm{a}+\mathrm{bx}}=\mathrm{t}\mathrm{give}\mathrm{a}+\mathrm{bx}={\mathrm{t}}^2\Rightarrow \mathrm{x}=\frac{{\mathrm{t}}^2-\mathrm{a}}{\mathrm{b}}(\mathrm{b}\ne 0)\Rightarrow$$$$\mathrm{I}=\int \frac{2\mathrm{t}}{\mathrm{b}\left(\frac{{\mathrm{t}}^2-\mathrm{a}}{\mathrm{b}}\right)\mathrm{t}}\mathrm{dt}=\int \frac{2\mathrm{dt}}{{\mathrm{t}}^2-\mathrm{a}}=\frac{1}{\sqrt{\mathrm{a}}}\int (\frac{1}{\mathrm{t}-\sqrt{\mathrm{a}}}-\frac{1}{\mathrm{t}+\sqrt{\mathrm{a}}})\mathrm{dt}=\frac{1}{\sqrt{\mathrm{a}}}\ln \mid \frac{\mathrm{t}-\sqrt{\mathrm{a}}}{\mathrm{t}+\sqrt{\mathrm{a}}}\mid +\mathrm{C}$$$$\mathrm{if}\mathrm{b}=0\Rightarrow \mathrm{I}=\frac{1}{\sqrt{\mathrm{a}}}\ln \mid \mathrm{x}\mid +\mathrm{c}$$$$$$

Commented by mathmax by abdo last updated on 06/Jun/20

$$\mathrm{t}=\sqrt{\mathrm{a}+\mathrm{bx}}\Rightarrow \mathrm{I}=\frac{1}{\sqrt{\mathrm{a}}}\ln \mid \frac{\sqrt{\mathrm{a}+\mathrm{bx}}-\sqrt{\mathrm{a}}}{\sqrt{\mathrm{a}+\mathrm{bx}}+\sqrt{\mathrm{a}}}\mid +\mathrm{C}$$

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