Question

How do you find the integral of $\ln ({x^2} - 1)dx$ ?

Answer 1

In order to determine the answer of above definite integral use the formula of integration by parts i.e. $\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }$ and assume $f(x) = {\cos ^{ - 1}}x$and$g'(x) = 1$ and calculate $f'(x)$and $g(x)$and put into the formula and use the substitution method to find the integral of the last term by splitting into two terms using partial fractions in the formula .

Formula:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C}$
$\int {\dfrac{1}{x} = \ln x + C}$
${\cos ^2}x + {\sin ^2}x = 1$
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }$

Complete step by step solution:
We are given an expression $\ln ({x^2} - 1)dx$ and we have to calculate its integral.
To calculate the integral of $\ln ({x^2} - 1)dx$ ,we will be using Integration by parts method

The formula for calculation of integration of parts is
$\int {f(x)g'(x)dx = f(x)g(x) - \int {f'(x)g(x)dx} }$
In our question Let assume
$f(x) = \ln ({x^2} - 1) And$g'(x) = 1$As we know that the derivative of function x is equal to 1 So$g(x) = x$and now calculating the derivative of$f(x)$with respect to x using rule of derivative$\dfrac{d}{{dx}}\ln X = \dfrac{1}{X}.\dfrac{d}{{dx}}X$By Chaining Rule therefore ,taking$X = {x^2} - 1$$
f'(x) = \dfrac{1}{{{x^2} - 1}}.\dfrac{d}{{dx}}({x^2} - 1) \\
= \dfrac{1}{{{x^2} - 1}}(2x) \\
= \dfrac{{2x}}{{{x^2} - 1}} \\
$now putting the values of$f(x),f'(x),g(x),and,g'(x)$into the formula of Integration by parts

$$\int {\ln ({x^2} - 1)dx = } x\ln ({x^2} - 1) - \int {x.\dfrac{{2x}}{{{x^2} - 1}}} dx \\\ = x\ln ({x^2} - 1) - \int {\dfrac{{2{x^2}}}{{{x^2} - 1}}dx} \\\ = x\ln ({x^2} - 1) - 2\int {\dfrac{{{x^2}}}{{{x^2} - 1}}dx} \\\ = x\ln ({x^2} - 1) - 2\int {\dfrac{{{x^2} - 1 + 1}}{{{x^2} - 1}}} dx \\\ = x\ln ({x^2} - 1) - 2\int {1 + \dfrac{1}{{{x^2} - 1}}} dx \\\ = x\ln ({x^2} - 1) - 2x - 2\int {\dfrac{1}{{{x^2} - 1}}} dx \\\$$

Using identity $({A^2} - {B^2}) = (A + B)(A - B)$
$= x\ln ({x^2} - 1) - 2x - 2\int {\dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} dx$
Using Partial fractions method to split $\dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}$ in to $\dfrac{1}{2}\left( {\dfrac{1}{{\left( {x - 1} \right)}} - \dfrac{1}{{\left( {x + 1} \right)}}} \right)$

= x\ln ({x^2} - 1) - 2x - 2\int {\dfrac{1}{2}\left( {\dfrac{1}{{\left( {x - 1} \right)}} - \dfrac{1}{{\left( {x + 1} \right)}}} \right)} dx \\\ = x\ln ({x^2} - 1) - 2x - \int {\dfrac{1}{{x - 1}} - \dfrac{1}{{x + 1}}} dx \\\ = x\ln ({x^2} - 1) - 2x - \left\\{ {\int {\dfrac{1}{{x - 1}}dx - \int {\dfrac{1}{{x + 1}}dx} } } \right\\} \\\ = x\ln ({x^2} - 1) - 2x - \left\\{ {\ln (x - 1) - \ln (x + 1)} \right\\} + C \\\ = x\ln ({x^2} - 1) - 2x - \ln (x - 1) + \ln (x + 1) + C \\\ \int {\ln ({x^2} - 1)dx = } x\ln ({x^2} - 1) - 2x + \dfrac{{\ln (x + 1)}}{{\ln (x + 1)}} + C \\\

Therefore, the integral $\int {\ln ({x^2} - 1)dx}$ is equal to $x\ln ({x^2} - 1) - 2x + \dfrac{{\ln (x + 1)}}{{\ln (x + 1)}} + C$ .

Note: 1.Use standard formula carefully while evaluating the integrals.
2. Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
The symbol $\int {f(x)dx}$ is read as the indefinite integral of $f(x)$with respect to x.