Mathematics

Evaluate $$\int {\dfrac{a} {b +
c{e^x}}}dx $$


SOLUTION

Let,

  $$ y=\int{\dfrac{a}{b+c{{e}^{x}}}}dx $$

 $$ put $$

 $$ t=b+c{{e}^{x}}\,\,\,\,\,\,\,\,\,\Rightarrow {{e}^{x}}=\dfrac{t-b}{c} $$

 $$ dt=c.{{e}^{x}}dx $$

 $$ dx=\dfrac{dt}{c.{{e}^{x}}} $$

 $$ Now, $$

 $$ y=\int{\dfrac{a}{t}}\dfrac{dt}{c.{{e}^{x}}} $$

 $$ y=\int{\dfrac{a}{t}}\dfrac{dt}{c.\left( \dfrac{t-b}{c} \right)} $$

 $$ y=a\int{\dfrac{1}{t\left( t-b \right)}}dt $$

 $$ y=a\int{\dfrac{1}{t-b}}-\dfrac{1}{t}dt $$

 $$ y=a\left[ \int{\dfrac{1}{t-b}}dt-\int{\dfrac{1}{t}dt} \right] $$

 $$ y=a\left[ \log \left( t-b \right)-\log t \right]+C $$

 $$ y=a\left[ \log \left( b+c{{e}^{x}}-b \right)-\log \left( b+c{{e}^{x}} \right) \right]+C $$

 $$ y=a.\log \dfrac{b+c{{e}^{x}}-b}{b+c{{e}^{x}}}+C $$

 $$ y=a.\log \dfrac{c.{{e}^{x}}}{b+c.{{e}^{x}}}+C $$

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