Calculate: F(x)=integral from 1 to ln(x) of 1+t t

Expression: $F\left( x \right)=\int_{ 1 }^{ \ln\left({x}\right) } 1+t \mathrm{d} t$

To simplify the function, evaluate the definite integral

$\int_{ 1 }^{ \ln\left({x}\right) } 1+t \mathrm{d} t$

To evaluate the definite integral first evaluate the indefinite integral

$\int{ 1+t } \mathrm{d} t$

Use the property of integral $\int{ f\left( x \right)\pmg\left( x \right) } \mathrm{d} x=\int{ f\left( x \right) } \mathrm{d} x\pm\int{ g\left( x \right) } \mathrm{d} x$

$\int{ 1 } \mathrm{d} t+\int{ t } \mathrm{d} t$

Use $\int{ 1 } \mathrm{d} x=x$ to evaluate the integral

$t+\int{ t } \mathrm{d} t$

Use $\int{ x } \mathrm{d} x=\frac{ {x}^{2} }{ 2 }$ to evaluate the integral

$t+\frac{ {t}^{2} }{ 2 }$

To evaluate the definite integral, return the limits of integration

$\left. \left( t+\frac{ {t}^{2} }{ 2 } \right) \right|_{ 1 }^{ \ln\left({x}\right) }$

Use $\left. F\left( x \right) \right|_{ a }^{ b }=F\left( b \right)-F\left( a \right)$ to evaluate the expression

$\ln\left({x}\right)+\frac{ {\ln\left({x}\right)}^{2} }{ 2 }-\left( 1+\frac{ {1}^{2} }{ 2 } \right)$

Simplify the expression

$\ln\left({x}\right)+\frac{ {\ln\left({x}\right)}^{2}-3 }{ 2 }$

The simplified function is

$F\left( x \right)=\ln\left({x}\right)+\frac{ {\ln\left({x}\right)}^{2}-3 }{ 2 }$