Question

How do you find the integral of $\ln \left( {{x^2} + 4} \right)$ ?

Answer 1

Integration is the process of finding the antiderivative. The integration of $g'\left( x \right)$ with respect to dx is given by $\int {g'\left( x \right)dx = g\left( x \right) + C}$, where C is the constant of integration and we can find the integral of the given function by using Integration by parts method.

Complete step by step answer:
The given function is $\ln \left( {{x^2} + 4} \right)$.As we need to find the integral, let us rewrite the function as,
$I = \int {\ln \left( {{x^2} + 4} \right)dx}$ ………………… 1
Apply rule of Integration by Parts as
$\int {u \cdot v \cdot dx} = u\int {v \cdot dx} - \int {\left( {\dfrac{{du}}{{dx}}\int {v \cdot dx} } \right)dx}$
Let,
$u = \ln \left( {{x^2} + 4} \right)$
Differentiate u with respect to x we get
$\dfrac{{du}}{{dx}} = \dfrac{1}{{{x^2} + 4}} \cdot 2x$
$\Rightarrow du = \dfrac{{2x}}{{{x^2} + 4}} \cdot dx$
Integrating v as:
$\int {v \cdot dx = x}$
Implies that,
$dv = dx \Rightarrow v = x$
i.e., $v = 1$
Using equation 1, let us find the integration
$I = \int {\ln \left( {{x^2} + 4} \right)dx}$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - \int {\left( {\dfrac{{2x}}{{{x^2} + 4}} \cdot x} \right)} dx$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - 2\int {\dfrac{{{x^2}}}{{{x^2} + 4}}} dx$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - 2\int {\dfrac{{\left( {{x^2} + 4} \right) - 4}}{{{x^2} + 4}}} dx$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - 2\int {\left( {\dfrac{{{x^2} + 4}}{{{x^2} + 4}} - \dfrac{4}{{{x^2} + 4}}} \right)} dx$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - 2\left( {\int {dx - 4\int {\dfrac{1}{{{x^2} + {2^2}}}dx} } } \right)$
Now find the integration of dx terms we get
$I = x\ln \left( {{x^2} + 4} \right) - 2\left( {x - 4 \cdot \dfrac{1}{2}\arctan \left( {\dfrac{x}{2}} \right)} \right)$
$\Rightarrow I = x\ln \left( {{x^2} + 4} \right) - 2x + 4\arctan \left( {\dfrac{x}{2}} \right) + C$
Therefore, the integral of $\ln \left( {{x^2} + 4} \right)$is
$\therefore I = x\ln \left( {{x^2} + 4} \right) - 2x + 4\arctan \left( {\dfrac{x}{2}} \right) + C$

Hence, the integral of $\ln \left( {{x^2} + 4} \right)$ is $x\ln \left( {{x^2} + 4} \right) - 2x + 4\arctan \left( {\dfrac{x}{2}} \right) + C$.

Note: There are different integration methods that are used to find an integral of some function, which is easier to evaluate the original integral. Hence, based on the function given we can find the integration of the function i.e., by using the integration methods as the details are given as additional information.