#10; #10;Let, #10; #10; #160; y=∫ab+ce^xdx #10; #10; #160; put #10; #10; #160; t=b+ce^x ⇒ #10;e^x=t bc #10 ...

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Integration by Substitution

Evaluate ∫a...

Question

Evaluate $\int \frac{a}{b + c e^{x}} d x$

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Solution

Let,

$y = \int \frac{a}{b + c e^{x}} d x$

$p u t$

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

$d t = c . e^{x} d x$

$d x = \frac{d t}{c . e^{x}}$

$N o w,$

$y = \int \frac{a}{t} \frac{d t}{c . e^{x}}$

$y = \int \frac{a}{t} \frac{d t}{c . (\frac{t - b}{c})}$

$y = a \int \frac{1}{t (t - b)} d t$

$y = a \int \frac{1}{t - b} - \frac{1}{t} d t$

$y = a [\int \frac{1}{t - b} d t - \int \frac{1}{t} d t]$

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

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

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

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

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