I have to compute this integral ∫0∞dt1+t4

Question

I have to compute this integral  ∫0∞dt1+t4

Alex Sheppard · Accepted Answer

Note that the substitution t=1u changes the integral to  ∫0∞u21+u4du  Split the integral into into two parts, 0 to 1, and 1 to infinity. On the second part, let t=1u. We get ∫01u21+u4du. Now u has done its duty, and is discarded for the more popular t. Our original integral is equal to  ∫011+t21+t4dt  There is now a minor miracle.  11−2t+t2+11+2t+t2=2(1+t2)1+t4  Complete the square(s) as usual.

karton · Accepted Answer

Let the considered integral be I i.e I=∫0∞11+t4dt Under the transformation t↦1/t, the integral is: I=∫0∞t21+t4dt⇒2I=∫0∞1+t21+t4dt=∫0∞1+1t2t2+1t2dt 2I=∫0∞1+1t2(t−1t)2+2dt Next, use the substitution 1−t/t=u⇒(1+1/t2)dt=du to get: 2I=∫−∞∞duu2+2⇒I=∫0∞duu2+2=π22

user_27qwe · Accepted Answer

If you don&#039;t want to do a partial fraction decomposition, put I:=∫0+∞dt1+t4. By the substitution x=1t on [0,+∞) and (−∞,0], we get ∫−∞+∞x21+x4dx. Now we have 4I=∫−∞+∞t2−2t+1(t2−2t+1)(t2+2t+1)dt=∫−∞+∞1t2+2t+1dt=∫−∞+∞1(t+22)2−12+1dt∫−∞+∞duu2+12=2∫−∞+∞du(2u)2+1=22arctan⁡(2u)|u=−∞u=+∞=2π2 and finally I=2π2

I have to compute this integral \int_0^{\infty}\frac{dt}{1+t^4}

Answered question

Answer & Explanation

New Questions in Calculus 1