Mathematics

Evaluate: $$\displaystyle \int x\log(1+x) dx$$


ANSWER

$$\dfrac{(2x^2 -2)\ln(x+1)-x^2 +2x}{4}+c$$


SOLUTION
$$\int { x } \log { \left( 1+x \right)  } dx$$

$$=\log { \left( 1+x \right)  } \int { x } dx-\int { \left[ \dfrac { d }{ dx } \log { \left( 1+x \right)  } .\int { x.dx }  \right]  } $$

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

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

$$=\dfrac{x^2}{2}\log \left(x+1\right) -\dfrac{1}{2}\left[ \int { \dfrac { \left( x-1 \right) \left( x+1 \right)  }{ \left( x+1 \right)  }  } dx+\int { \dfrac { 1 }{ \left( x+1 \right)  } dx }  \right] $$

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

$$=\dfrac{1}{2}\log \left( x+1\right)\left(x^2-1\right)-\dfrac{1}{2}\left[ \dfrac { { x }^{ 2 } }{ 2 } -x \right] +c.$$

 $$=\dfrac{1}{2}\log \left( x+1\right)\left(x^2-1\right)-\dfrac{1}{2}\left[ \dfrac { { x }^{ 2 } }{ 2 } -x \right] +c.$$

Hence, the answer is $$\dfrac{(2x^2 -2)\ln(x+1)-x^2 +2x}{4}+c$$
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Single Correct Medium Published on 17th 09, 2020
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