Question

How do you integrate $\dfrac{1}{2x+1}$?

Answer 1

In calculus, an integral is used to find the area under the graph of an equation. Integration, the process of finding an integral, is the reverse of differentiation, the process of finding a derivative. Integration of the inverse of a polynomial expression of degree one is equal to natural logarithm of modulus of the polynomial divided by the derivative of the polynomial. That is, If P(x) is the polynomial of degree one then, $\int{\dfrac{1}{P(x)}dx=\dfrac{1}{\left( \dfrac{d(P(x))}{dx} \right)}\ln |P(x)|}+C(constant)$.

Complete step by step answer:
As per the given question we need to integrate $\dfrac{1}{2x+1}$. Here, we have a polynomial expression of degree one (2x+1) in the denominator. So, when we integrate $\dfrac{1}{2x+1}$, we get the natural logarithm of the modulus of the polynomial expression (2x+1) divided by the derivative of the polynomial (2x+1). That is, we can express it as
$\int{\dfrac{1}{P(x)}dx=\dfrac{1}{\left( \dfrac{d(P(x))}{dx} \right)}\ln |P(x)|}+C(constant)$ ---------(1)
As we know that, for a general polynomial expression like $P(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+......+{{a}_{n}}{{x}^{n}}={{a}_{0}}{{x}^{0}}+{{a}_{1}}{{x}^{1}}+{{a}_{2}}{{x}^{2}}+......+{{a}_{n}}{{x}^{n}}$, the derivative of this polynomial expression can be written as
$d\left( P(x) \right)=0+{{a}_{1}}{{x}^{1-1}}+{{a}_{2}}{{x}^{2-1}}+......+{{a}_{n}}{{x}^{n-1}}={{a}_{1}}+{{a}_{2}}x+{{a}_{3}}{{x}^{2}}+......+{{a}_{n}}{{x}^{n-1}}.$
Here, we have the polynomial expression of degree one, P(x) = (2x+1). Then we get the derivative of the polynomial expression P(x) from below equation:
$d\left( P(x) \right)=d(2x+1)=(2{{x}^{1-1}}+0)dx=2dx$
$\therefore \dfrac{d(P(x))}{dx}=2$ --------(2)
Now, on substituting the obtained derivative from equation (2) into the equation (1), we get
$\int{\dfrac{1}{P(x)}dx=\dfrac{1}{2}\ln |P(x)|}+C(constant)$
(we write constant C as we don’t know the bounds).

Therefore, $\dfrac{1}{2}\ln |2x+1|+C$ is the expression obtained by integrating $\dfrac{1}{2x+1}$.

Note: Basically, the integration takes place by substitution method. Here, we change the $dx\to d(2x+1)$ form since we have (2x+1) in the denominator of the given expression. As we know that, $d(2x+1)$ derivative value equals $2dx$, we replace $dx$ by $d(2x+1)$. A common error made in such a type of problem is ignoring the constant term C after integration.