Question

Evaluate the following:
$\int \frac{e^{3x}}{e^{3x}+1}.dx$

• A. $\frac{1}{3}ln(e^{3x}-1)+C$
• B. $\frac{1}{3}\ln (e^{4x}+1)+C$
• C. $\frac{1}{3}ln(e^{3x}+1)+C$
• D. None of these

Answer 1

The correct option is C $\frac{1}{3}ln(e^{3x}+1)+C$

Let , $e^{3x}+1=t$

So, $3.e^{3x}.dx=dt$

On substituting values , we get

$\int \frac{dt}{3t}$

$⟹\frac{1}{3}lnt+C$

$⟹\frac{1}{3}ln(e^{3x}+1)+C$