Question

g(x)=`int_sqrtx^xtan^(-1)tdt` Find? 1) g(`sqrt3`) 2) g'(`sqrt3`) 3) g"(`sqrt3`)

Answer 1

Please see below.

Explanation:

Here,

`g(x)=int_sqrtx^x tan^-1tdt=int_sqrtx^x(1) tan^-1tdt`

`"Using "color(blue) "Integration by Parts:"`

`I=tan^-1t int1dt-int(d/(dt)(tan^-1t)int1dt)dt`

`I=[tan^-1txxt]_sqrtx^x -int_sqrtx^xt/(1+t^2)dt`

`=[t*tan^-1t]_sqrtx^x-1/2int_sqrtx^x(2t)/(1+t^2)dt`

`=xtan^-1x-sqrtxtan^-1sqrtx-1/2[ln|1+t^2|]_sqrtx^x`

`g(x)=xtan^-1x-sqrtxtan^-1sqrtx-1/2[ln|1+x^2|-ln|1+x|]`

`g(x)=xtan^-1x-sqrtxtan^-1sqrtx-1/2ln|(1+x^2)/(1+x)|...to(A)`

#:.g(sqrt3)=sqrt3xxpi/3-root(4)3tan^-1root(4)3-1/2ln| (1+sqrt3)/(1+root(4)3)|#

From `(A)`,

`g'(x)=tan^-1x-1/(2sqrtx)tan^-1sqrtx`

`g'(sqrt3)=pi/3-1/(2root(4)3)tan^-1root(4)3`

g"`(x)=1/(1+x^2)-1/(4x(1+x))+1/(4x^(3/2))tan^-1sqrtx`

g"`(sqrt3)=1/4-1/(4sqrt3(1+sqrt3))+1/(4(3)^(3/4))tan^-1root(4)3`