Question #0053d

2 Answers
Dec 7, 2017

#sqrt2/4*arctan((t^2-1)/(tsqrt2))+sqrt2/8Ln((t^2-tsqrt2+1)/(t^2+tsqrt2+1))+C#

Explanation:

#int t^2/(t^4+1)*dt#

=#int (dt)/(t^2+1/t^2)*dt#

=#1/2int (2dt)/(t^2+1/t^2)*dt#

=#1/2int ((1+1/t^2)*dt)/(t^2+1/t^2)#+#1/2int ((1-1/t^2)*dt)/(t^2+1/t^2)#

#A=1/2int ((1+1/t^2)*dt)/(t^2+1/t^2)#

=#1/2int (d(t-1/t))/[(t-1/t)^2+2]#

=#sqrt2/4*arctan((t-1/t)/sqrt2)+C#

=#sqrt2/4*arctan((t^2-1)/(tsqrt2))+C#

#B=1/2int ((1-1/t^2)*dt)/(t^2+1/t^2)#

#B=1/2int (d(t+1/t))/[(t+1/t)^2-2]#

=#sqrt2/8Ln(t+1/t-sqrt2)-sqrt2/8Ln(t+1/t+sqrt2)#

=#sqrt2/8Ln((t+1/t-sqrt2)/(t+1/t+sqrt2))#

=#sqrt2/8Ln((t^2-tsqrt2+1)/(t^2+tsqrt2+1))#

Thus,

#int t^2/(t^4+1)*dt#

=#A+B#

=#sqrt2/4*arctan((t^2-1)/(tsqrt2))+sqrt2/8Ln((t^2-tsqrt2+1)/(t^2+tsqrt2+1))+C#

Dec 7, 2017

#1/(2sqrt2)arc tan{1/sqrt2(t-1/t)}+1/(4sqrt2)ln|(t^2-sqrt2t+1)/(t^2+sqrt2t+1)|+C.#

Explanation:

Let, #I=intt^2/(1+t^4)dt=1/2int(2t^2)/(1+t^4)dt,#

#=1/2int{(t^2+1)+(t^2-1)}/(1+t^4)dt,#

#=1/2int(t^2+1)/(1+t^4)dt+1/2int(t^2-1)/(1+t^4)dt,#

#=1/2I_1+1/2I_2...........(star),# where,

#I_1=int(t^2+1)/(1+t^4)dt,#

#=int{t^2(1+1/t^2)}/{t^2(t^2+1/t^2)}dt,#

#=int(1+1/t^2)/(t^2+1/t^2)dt.#

Now, we substitute,

#(t-1/t)=u," so that, "(1+1/t^2)dt=du.#

Also, #t^2+1/t^2=(t-1/t)^2+2.#

#:. I_1=int1/(u^2+2)du=1/sqrt2arc tan(u/sqrt2), i.e.,#

# I_1=1/sqrt2arc tan{1/sqrt2(t-1/t)},#

# rArr I_1=1/sqrt2arc tan((t^2-1)/(sqrt2t))........(star_1).#

Similarly, #I_2=int(1-1/t^2)/(t^2+1/t^2)dt,# becomes, on using the

substitution #(t+1/t)=v, I_2=int1/(v^2-2)dv,#

#=1/(2sqrt2)ln|(v-sqrt2)/(v+sqrt2)|,#

#rArr I_2=1/(2sqrt2)ln|(t^2-sqrt2t+1)/(t^2+sqrt2t+1)|.......(star_2).#

Altogether, from #(star),(star_1) and (star_2),# we have,

#I=1/(2sqrt2)arc tan{1/sqrt2(t-1/t)}#

#+1/(4sqrt2)ln|(t^2-sqrt2t+1)/(t^2+sqrt2t+1)|+C.#

Enjoy Maths.!