Question

Question #0053d

Answer 1

`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`

Answer 2

`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.!