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Question: Evaluate the value of \(\int {x\log \left( {1 + x} \right)dx} \) A. \(\dfrac{{\left( {2{x^2} + 2}...

Evaluate the value of ∫xlog⁡(1+x)dx\int {x\log \left( {1 + x} \right)dx}
A. (2x2+2)ln⁡(x+1)−x2+2x4+C\dfrac{{\left( {2{x^2} + 2} \right)\ln \left( {x + 1} \right) - {x^2} + 2x}}{4} + C
B. (2x2+2)ln⁡(x+1)−x2−2x2+C\dfrac{{\left( {2{x^2} + 2} \right)\ln \left( {x + 1} \right) - {x^2} - 2x}}{2} + C
C. (2x2−2)ln⁡(x+1)−x2+2x4+C\dfrac{{\left( {2{x^2} - 2} \right)\ln \left( {x + 1} \right) - {x^2} + 2x}}{4} + C
D. (2x2−2)ln⁡(x+1)−x2+4x2+C\dfrac{{\left( {2{x^2} - 2} \right)\ln \left( {x + 1} \right) - {x^2} + 4x}}{2} + C

Explanation

Solution

Firstly, let I=∫xlog⁡(1+x)dxI = \int {x\log \left( {1 + x} \right)dx} .
Then, find the value of integral using the formula ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\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(x)=log⁡(1+x)f\left( x \right) = \log \left( {1 + x} \right) and g(x)=xg\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 ∫xlog⁡(1+x)dx\int {x\log \left( {1 + x} \right)dx} .
Let I=∫xlog⁡(1+x)dxI = \int {x\log \left( {1 + x} \right)dx}
Since, we know that the integral of the product of two functions f(x)f\left( x \right) and g(x)g\left( x \right) is ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\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(x)=log⁡(1+x)f\left( x \right) = \log \left( {1 + x} \right) and g(x)=xg\left( x \right) = x .
⇒I=log⁡(1+x)⋅∫xdx−∫[ddxlog⁡(1+x)⋅∫xdx]dx\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}
⇒I=x22log⁡(1+x)−∫11+x×x22dx\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \int {\dfrac{1}{{1 + x}} \times \dfrac{{{x^2}}}{2}dx}
⇒I=x22log⁡(1+x)−12∫x2x+1dx\Rightarrow I = \dfrac{{{x^2}}}{2}\log \left( {1 + x} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{{x + 1}}dx} … (1)
Let, ∫x2x+1dx=I1\int {\dfrac{{{x^2}}}{{x + 1}}dx} = {I_1}
⇒I1=∫x2x+1dx\Rightarrow {I_1} = \int {\dfrac{{{x^2}}}{{x + 1}}dx}
=∫x2−1+1x+1dx =∫x2−1x+1dx+∫1x+1dx =∫(x+1)(x−1)x+1dx+log⁡(1+x) =∫(x−1)dx+log⁡(1+x) =x22−x+log⁡(1+x)+C  = \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, I1=x22−x+log⁡(1+x)+C{I_1} = \dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right) + C .
Now, substitute the value of I1=x22−x+log⁡(1+x){I_1} = \dfrac{{{x^2}}}{2} - x + \log \left( {1 + x} \right) in equation (1).
⇒I=x22log⁡(1+x)−12[x22−x+log⁡(1+x)]+C\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
⇒I=x22log⁡(1+x)−12(x22−x)−12log⁡(1+x)+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
⇒I=(x2−1)×12log⁡(1+x)−12(x2−2x2)+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
⇒I=(x2−12)log⁡(1+x)−(x2−2x4)+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
⇒I=2(x2−1)log⁡(1+x)−(x2−2x)4+C\Rightarrow I = \dfrac{{2\left( {{x^2} - 1} \right)\log \left( {1 + x} \right) - \left( {{x^2} - 2x} \right)}}{4} + C
⇒I=(2x2−2)log⁡(1+x)−x2+2x4+C\Rightarrow I = \dfrac{{\left( {2{x^2} - 2} \right)\log \left( {1 + x} \right) - {x^2} + 2x}}{4} + C
Thus, we get ⇒∫xlog⁡(1+x)dx=(2x2−2)log⁡(1+x)−x2+2x4+C\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 ∫f(x)g(x)dx=f(x)∫g(x)dx−∫[ddxf(x)⋅∫g(x)dx]dx\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(x)f\left( x \right) and g(x)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(x)f\left( x \right) must be a function lying before the function g(x)g\left( x \right) in the ILATE rule.
In the given question, we chose f(x)=log⁡(1+x)f\left( x \right) = \log \left( {1 + x} \right) and g(x)=xg\left( x \right) = x , because log⁡(1+x)\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.