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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int\sqrt{\sec x-1}dx=$
• A. $\displaystyle 2\log\left(\cos{\frac{x}{2}+}\sqrt{\cos^{2}\frac{x}{2}-\frac{1}{2}}\right)+C$
• B. $\displaystyle \log\left(\cos{\frac{x}{2}+}\sqrt{\cos^{2}\frac{x}{2}-\frac{1}{2}}\right)+C$
• C. $\displaystyle 2\log\left(\cos\frac{x}{2}-\sqrt{\cos^{2}\frac{x}{2}-\frac{1}{2}}\right)+c$
• D. $\displaystyle -2\log\left(\cos{\frac{x}{2}+}\sqrt{\cos^{2}\frac{x}{2}-\frac{1}{2}}\right)+C$

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Q2 Single Correct Medium
$\int_0^\infty {{x^n}{e^{ - x}}dx}$ (n is +ve integer) is equal to
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• D. $n!$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \frac{x^{4}}{(x-1)(x^{2}+1)}dx$
• A. $\displaystyle x^{2}+x+\frac{1}{2}log(x-1)-\frac{1}{4}log(x^{2}+1)-\frac{1}{2}\tan ^{-1}x+c$
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• C. $\displaystyle \frac{1}{2}x^{2}+\frac{1}{2}log(x-1)-\frac{1}{4}log(x^{2}+1)-\frac{1}{2}\tan ^{-1}x+c$
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Q4 Subjective Medium
$\displaystyle \int \dfrac {x^{4}}{x^{2}+1}dx$

Let $\displaystyle 2I_{1}+I_{2}=\int \frac {e^{x}}{e^{2x}+e^{-2x}}dx$  and  $\displaystyle I_{1}+2I_{2}=\int \frac {e^{-x}}{e^{2x}+e^{-2x}}dx$