Question

Evaluate the value of $\int {x\log \left( {1 + x} \right)dx} $
A. $\dfrac{{\left( {2{x^2} + 2} \right)\ln \left( {x + 1} \right) - {x^2} + 2x}}{4} + C$
B. $\dfrac{{\left( {2{x^2} + 2} \right)\ln \left( {x + 1} \right) - {x^2} - 2x}}{2} + C$
C. $\dfrac{{\left( {2{x^2} - 2} \right)\ln \left( {x + 1} \right) - {x^2} + 2x}}{4} + C$
D. $\dfrac{{\left( {2{x^2} - 2} \right)\ln \left( {x + 1} \right) - {x^2} + 4x}}{2} + C$

Answer 1

Hint: Firstly, let $I = \int {x\log \left( {1 + x} \right)dx} $ .
Then, find the value of integral using the formula $\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} $ , here $f\left( x \right) = \log \left( {1 + x} \right)$ and $g\left( x \right) = x$ .
Thus, solve it further and choose the correct answer.

Complete step-by-step answer:
Here, we have to find the value of $\int {x\log \left( {1 + x} \right)dx} $ .
Let $I = \int {x\log \left( {1 + x} \right)dx} $
Since, we know that the integral of the product of two functions \[f\left( x \right)\] and \[g\left( x \right)\] is $\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} $ .
So, we will find the integral of I, by using the above formula, where $f\left( x \right) = \log \left( {1 + x} \right)$ and $g\left( x \right) = x$ .
 \[\Rightarrow I = \log \left( {1 + x} \right) \cdot \int {xdx} - \int {\left[ {\dfrac{d}{{dx}}\log \left( {1 + x} \right) \cdot \int {xdx} } \right]dx} \]
 $\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \int {\dfrac{1}{{1 + x}} \times \dfrac{{{x^2}}}{2}dx} $
 $\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx} $ … (1)
Let, $\int {\dfrac{{{x^2}}}{{x + 1}}dx} = {I_1}$
 $\Rightarrow {I_1} = \int {\dfrac{{{x^2}}}{{x + 1}}dx} $
 $
   = \int {\dfrac{{{x^2} - 1 + 1}}{{x + 1}}dx} \\
   = \int {\dfrac{{{x^2} - 1}}{{x + 1}}dx} + \int {\dfrac{1}{{x + 1}}dx} \\
   = \int {\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{x + 1}}dx} + \log \left( {1 + x} \right) \\
   = \int {\left( {x - 1} \right)dx} + \log \left( {1 + x} \right) \\
   = \dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right) + C \\
 $
Thus, ${I_1} = \dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right) + C$ .
Now, substitute the value of ${I_1} = \dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right)$ in equation (1).
 $\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \dfrac{1}{2}\left[ {\dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right)} \right] + C$
 $\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2} - x} \right) - \dfrac{1}{2}\log \left( {1 + x} \right) + C$
 $\Rightarrow I = \left( {{x^2} - 1} \right) \times \dfrac{1}{2}\log \left( {1 + x} \right) - \dfrac{1}{2}\left( {\dfrac{{{x^2} - 2x}}{2}} \right) + C$
 $\Rightarrow I = \left( {\dfrac{{{x^2} - 1}}{2}} \right)\log \left( {1 + x} \right) - \left( {\dfrac{{{x^2} - 2x}}{4}} \right) + C$
Now, taking LCM
 \[\Rightarrow I = \dfrac{{2\left( {{x^2} - 1} \right)\log \left( {1 + x} \right) - \left( {{x^2} - 2x} \right)}}{4} + C\]
 $\Rightarrow I = \dfrac{{\left( {2{x^2} - 2} \right)\log \left( {1 + x} \right) - {x^2} + 2x}}{4} + C$
Thus, we get $\Rightarrow \int {x\log \left( {1 + x} \right)dx} = \dfrac{{\left( {2{x^2} - 2} \right)\log \left( {1 + x} \right) - {x^2} + 2x}}{4} + C$ .
So, option (C) is correct.

Note: The formula for integration by parts is given by $\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \int {\left[ {\dfrac{d}{{dx}}f\left( x \right) \cdot \int {g\left( x \right)dx} } \right]dx} $ .
In the formula for integration by parts, \[f\left( x \right)\] and \[g\left( x \right)\] are chosen by the method of ILATE.
ILATE is known as Inverse trigonometric functions, Logarithm functions, Algebraic functions, Trigonometric functions and Exponential functions.
The function \[f\left( x \right)\] must be a function lying before the function \[g\left( x \right)\] in the ILATE rule.
In the given question, we chose $f\left( x \right) = \log \left( {1 + x} \right)$ and $g\left( x \right) = x$ , because $\log \left( {1 + x} \right)$ is a logarithm function and x is an algebraic function and in ILATE rule logarithm functions lie before algebraic functions.