∫ x ln ( x ) d x {\displaystyle \int x\ln({x})dx} u = x + 1 x = u − 1 d u = d x {\displaystyle {\begin{aligned}u&=x+1\\[2ex]x&=u-1\\[2ex]du&=dx\\[2ex]\end{aligned}}} ∫ ( u − 1 ) ln ( u ) d u {\displaystyle \int (u-1)\ln({u})du} w = ln u d v = ( u − 1 ) d v {\displaystyle w=\ln {u}\qquad dv=(u-1)dv} d u = 1 u d u v = 1 2 u 2 − u {\displaystyle du={\frac {1}{u}}du\qquad v={\frac {1}{2}}u^{2}-u} ∫ ( u − 1 ) ln ( u ) d u = ln u ⋅ 1 2 u 2 − u − ∫ ( 1 2 u 2 − u ) ⋅ 1 u d u = ln u ⋅ 1 2 u 2 − u − ∫ ( 1 2 u − 1 ) d u = ln u ⋅ 1 2 u 2 − u − [ 1 4 u 2 − u ] + c = ln ( 1 + x ) ( 1 2 ( x 2 + 2 x + 1 ) − ( 1 + x ) ) − ( 1 4 ( 1 + x ) 2 − ( 1 + x ) ) + c = ln ( 1 + x ) ( 1 2 x 2 + x + 1 2 − x − 1 ) − ( 1 4 ( 1 + x ) 2 − ( 1 + x ) ) + c = ln ( 1 + x ) ( 1 2 x 2 − 1 2 ) − 1 4 x 2 − 1 2 x − 1 4 + 1 + x + c = ln ( 1 + x ) ( 1 2 x 2 − 1 2 ) − 1 4 x 2 + 1 2 x + 3 4 + c {\displaystyle {\begin{aligned}\int (u-1)\ln({u})du&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u^{2}-u)\cdot {\frac {1}{u}}du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u-1)du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-[{\frac {1}{4}}u^{2}-u]+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}(x^{2}+2x+1)-(1+x))-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}x^{2}+x+{\frac {1}{2}}-x-1)-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}-{\frac {1}{2}}x-{\frac {1}{4}}+1+x+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}+{\frac {1}{2}}x+{\frac {3}{4}}+c\\[2ex]\end{aligned}}}
![{\displaystyle {\begin{aligned}u&=x+1\\[2ex]x&=u-1\\[2ex]du&=dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/384447aed379c8029bdd9f79fd82b9bec930cc01)
![{\displaystyle {\begin{aligned}\int (u-1)\ln({u})du&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u^{2}-u)\cdot {\frac {1}{u}}du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-\int ({\frac {1}{2}}u-1)du\\[2ex]&=\ln {u}\cdot {\frac {1}{2}}u^{2}-u-[{\frac {1}{4}}u^{2}-u]+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}(x^{2}+2x+1)-(1+x))-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln({1+x})({\frac {1}{2}}x^{2}+x+{\frac {1}{2}}-x-1)-({\frac {1}{4}}(1+x)^{2}-(1+x))+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}-{\frac {1}{2}}x-{\frac {1}{4}}+1+x+c\\[2ex]&=\ln {(1+x})({\frac {1}{2}}x^{2}-{\frac {1}{2}})-{\frac {1}{4}}x^{2}+{\frac {1}{2}}x+{\frac {3}{4}}+c\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/51c786398b51ec56ce6b7ce2e4b7385f4d4b4d8e)