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Find I = ∫ 1 2 x ln x d x {\displaystyle I=\int _{1}^{2}x\ln {x}\mathrm {d} x\!} .
Find I = ∫ ( x + ln x + e x ) d x {\displaystyle I=\int (x+\ln {x}+e^{x})\,\mathrm {d} x\!} .
I = ∫ x d x + ∫ ln x d x + ∫ e x d x {\displaystyle I=\int x\,\mathrm {d} x\!+\int \ln {x}\,\mathrm {d} x\!+\int e^{x}\,\mathrm {d} x\!}
I = ( x 2 2 + c 1 ) + ∫ ln x d x + ( e x + c 3 ) {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+\int \ln {x}\,\mathrm {d} x\!+(e^{x}+c_{3})}
Let u = ln x {\displaystyle u=\ln {x}} and d v = d x {\displaystyle dv=\mathrm {d} x\!} .
Then,
d u = 1 x d x {\displaystyle du={\frac {1}{x}}\mathrm {d} x\!} and v = x {\displaystyle v=x} .
So,
I = ( x 2 2 + c 1 ) + < m a t h > u v − ∫ v d u {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+<math>uv-\int v\,\mathrm {d} u\!} + (e^x +c_3)</math>
I = ( x 2 2 + c 1 ) + < m a t h > ( ln x ) ( x ) − ∫ v d u {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+<math>(\ln {x})(x)-\int v\,\mathrm {d} u\!} + (e^x +c_3)</math>
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and 
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and 
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+ (e^x +c_3)</math>
+ (e^x +c_3)</math>
