Find I = ∫ 1 2 x ln x d x {\displaystyle I=\int _{1}^{2}x\ln {x}\,\mathrm {d} x\!} .
Let u = ln x {\displaystyle u=\ln {x}} and d v = x d x {\displaystyle dv=x\,\mathrm {d} x\!} .
Then, d u = 1 x d x {\displaystyle du={\frac {1}{x}}\,\mathrm {d} x\!} and v = x 2 2 {\displaystyle v={\frac {x^{2}}{2}}} .
So,
Failed to parse (syntax error): {\displaystyle I=uv\right|_1^2-\int_1^2 v\,\mathrm du.\!}
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} .
I = ( x 2 2 + c 1 ) + u v − ∫ v d u + ( e x + c 3 ) {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+uv-\int v\,\mathrm {d} u\!+(e^{x}+c_{3})}
I = ( x 2 2 + c 1 ) + ( ln x ) ( x ) − ∫ ( x ) ( 1 x d x ) + ( e x + c 3 ) {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(\ln {x})(x)-\int (x)\,({\frac {1}{x}}\mathrm {d} x\!)+(e^{x}+c_{3})}
I = ( x 2 2 + c 1 ) + ( ln x ) ( x ) − ∫ d x + ( e x + c 3 ) {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(\ln {x})(x)-\int \mathrm {d} x\!+(e^{x}+c_{3})}
I = ( x 2 2 + c 1 ) + ( x ln x − x + c 2 ) + ( e x + c 3 ) {\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(x\ln {x}-x+c_{2})+(e^{x}+c_{3})}
I = x 2 2 + x ln x − x + e x + C {\displaystyle I={\frac {x^{2}}{2}}+x\ln {x}-x+e^{x}+C}
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