What is the antiderivative of (t^2)/(t^2 + 1) dt from 1 to sq. rt. x?

I got (x^2)/[(2)(x^2 + 1)(sq. rt. x)]. Is this correct?

1 Answer
Feb 2, 2018

# sqrtx-arc tansqrtx-1+pi/4#.

Explanation:

Let, #I=int_1^sqrtxt^2/(t^2+1)dt#.

#:. I=int_1^sqrtx{(t^2+1)-1}/(t^2+1)dt#,

#=int_1^sqrtx{(t^2+1)/(t^2+1)-1/(t^2+1)}dt#,

#=int_1^sqrtx{1-1/(t^2+1)}dt#,

#=[t-arc tant]_1^sqrtx#,

#=(sqrtx-arc tansqrtx)-(1-arc tan1)#,

#rArr I=sqrtx-arc tansqrtx-1+pi/4#.