Anyone think they can tackle this question?
Having a little bit of trouble.
I used this site:
and got this answer:
I haven't tried to make sense of it yet though.
Something Mathematical to do.
∞∫x=01(x+√1+x2)2 dxsubstitute: x=tan(z)dx=(1+tan2(z)) dz new limits: z=arctan(0)⇒z=0z=arctan(∞)⇒z=π2 =π2∫z=01+tan2(z)(tan(z)+√1+tan2(z))2 dz 1+tan2(z)=1cos2(z) =π2∫z=01cos2(z)⋅(11cos2(z)+tan2(z)+2⋅sin(z)cos2(z)) dz=π2∫z=011+2⋅sin(z)+sin2(z) dz=π2∫z=01(1+sin(z))2 dzsubstitute: t=tan(z2)dt=12⋅(1+tan2(z2)) dz=12⋅(1+t2) dz new limits: t=tan(02)⇒t=0t=tan(π22)⇒t=1 =1∫t=02 dt1+t2⋅(1(1+sin(z))2) sin(z)=2t1+t2 =1∫t=02 dt1+t2⋅(1(1+2t1+t2)2)=2⋅1∫t=01+t21+4t+6t2+4t3+t4 dt=2⋅1∫t=01+t2(1+t)4 dt
Partial fraction decomposition: 1+t2(1+t)4=A1+t+B(1+t)2+C(1+t)3+D(1+t)41+t2=A⋅(1+t)3+B⋅(1+t)2+C⋅(1+t)+D(1)t=−1:⇒2=D(2)t=0:⇒−1=A+B+C(3)t=1:⇒0=8A+4B+2C|:20=4A+2B+C(4)t=−2:⇒3=−A+B−C(2)+(4):−1+3=2B⇒B=1(3)+(4):0+3=3A+3B|:31=A+B|B=11=A+1⇒A=0(2):C=−1−A−B=−1−0−1⇒C=−2
=21∫t=0(1(1+t)2−2⋅1(1+t)3+2⋅1(1+t)4) dt=21∫t=01(1+t)2 dt−41∫t=01(1+t)3 dt+41∫t=01(1+t)4 dt=2[−11+t]1t=0−42[−1(1+t)2]1t=0+43[−1(1+t)3]1t=0=2[−12−(−1)]−42[−14−(−1)]+43[−18−(−1)]=2[−12+1]−2[−14+1]+43[−18+1]=2⋅12−2⋅34+43⋅78=1−32+76=66−96+76=136−96=46=23
∞∫x=01(x+√1+x2)2 dx=23