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

1 Answer
May 6, 2018

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