First slide

Methods of integration

Question

$If J = \int \sqrt{e^{2 x} + a e^{x}} d x then J is equal to$

Moderate

A

$\sqrt{e^{2 x} + a e^{x}} + a \log |e^{\frac{x}{2}} + \sqrt{e^{x} + a}| + C$
B

$\sqrt{e^{2 x} + a e^{x}} + a \log |e^{x} + a e^{2 x}| + C$
C

$\sqrt{e^{2 x} + a e^{x}} + a \log |e^{\frac{x}{2}} + e^{a x}| + C$
D

$\sqrt{e^{2 x} + a e^{x}} + \log |e^{x} + \sqrt{e^{x} + a}| + C$

Solution

$J = \int e^{x} \sqrt{1 + a e^{- x}} d x$
$Put e^{x} = t^{2}$
$\begin{array}{l} e^{x} d^{x} = 2 t d t \\ J = \int 2 t \sqrt{1 + \frac{a}{t^{2}}} d t \\ J = 2 \int \sqrt{t^{2} + a} d t \\ = 2 [\frac{t}{2} \sqrt{t^{2} + a} + \frac{a}{2} \log |t + \sqrt{t^{2} + a}|] + C \\ J = [\sqrt{e^{2 x} + a e^{x}} + a \log |e^{\frac{x}{2}} + \sqrt{e^{x} + a}|] + C \end{array}$

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