Jw116104

Find ${\displaystyle I=\int _{1}^{2}x\ln {x}\,\mathrm {d} x\!}$ .

Let ${\displaystyle u=\ln {x}}$  and ${\displaystyle dv=x\,\mathrm {d} x\!}$ .

Then, ${\displaystyle du={\frac {1}{x}}\,\mathrm {d} x\!}$  and ${\displaystyle v={\frac {x^{2}}{2}}}$ .

So,

Failed to parse (syntax error): {\displaystyle I=uv\right|_1^2-\int_1^2 v\,\mathrm du.\!}

Find ${\displaystyle I=\int (x+\ln {x}+e^{x})\,\mathrm {d} x\!}$ .

${\displaystyle I=\int x\,\mathrm {d} x\!+\int \ln {x}\,\mathrm {d} x\!+\int e^{x}\,\mathrm {d} x\!}$ 

${\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+\int \ln {x}\,\mathrm {d} x\!+(e^{x}+c_{3})}$ 

Let ${\displaystyle u=\ln {x}}$  and ${\displaystyle dv=\mathrm {d} x\!}$ .

Then, ${\displaystyle du={\frac {1}{x}}\mathrm {d} x\!}$  and ${\displaystyle v=x}$ .

So,

${\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+uv-\int v\,\mathrm {d} u\!+(e^{x}+c_{3})}$ 

${\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(\ln {x})(x)-\int (x)\,({\frac {1}{x}}\mathrm {d} x\!)+(e^{x}+c_{3})}$ 

${\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(\ln {x})(x)-\int \mathrm {d} x\!+(e^{x}+c_{3})}$ 

${\displaystyle I=({\frac {x^{2}}{2}}+c_{1})+(x\ln {x}-x+c_{2})+(e^{x}+c_{3})}$ 

${\displaystyle I={\frac {x^{2}}{2}}+x\ln {x}-x+e^{x}+C}$