Question

# Integrate the following with respect to x:$\int {\dfrac{1}{{2x + 3}}dx}$.

Hint: Simple substitution of the denominator to some variable will help simplifying the integral and reducing it to a standard integral. Use this technique to evaluate this integral.

Let $I = {\text{ }}\int {\dfrac{1}{{2x + 3}}dx}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)
Let 2x+3 = pâ€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)

Now, differentiate both the sides of equation (2) we get,
$\Rightarrow 2dx = dp$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

Make this substitution back into the main integral$I$, substituting (3) in equation (1) we get
$\Rightarrow I = {\text{ }}\dfrac{1}{2}\int {\dfrac{1}{p}dp}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (4)

Now, we know that the standard integral of,$\int {\dfrac{1}{x}dx = \log x}$ â€¦â€¦â€¦â€¦â€¦â€¦ (5)
So the value of equation (4) will be, using above equation (5) we get,
$I = \dfrac{1}{2}\log p + c$

Now, letâ€™s substitute the value of p back into the above integral we get
$I = \dfrac{1}{2}\log \left( {2x + 3} \right) + c$ Using equation (2)

Note: Whenever we face such problems always try and simplify the integral via method of substitution. This will help simplifying the integral into a standard from. Donâ€™t forget to substitute back the variable assumed and take the solution back to the main variable provided in question. The constant of integration is also to be taken care of in exams.