Q.

Let f(x)=∫dx(3+4x2)4-3x2,|x|<23. If f(0)=0 and f(1)=1αβtan-1(αβ), α,β>0, then α2+β2 is equal to _______ .        [2023]


Ans.

(28)

Given, f(x)=∫dx(3+4x2)4-3x2 

x=1t⇒dx=-1t2dt

⇒f(x)=∫-1t2dt(3t2+4)t24t2-3t 

=-∫-t dt(3t2+4)4t2-3  Put 4t2-3=z2⇒t dt=z4 dz

=-14∫z dz(3(z2+34)+4)z=∫-dz3z2+25=-13∫dzz2+(53)2

=∫-1335tan-1(3z5)+C=-153tan-1(354t2-3)+C

f(x)=-153tan-1(354-3x2x2)+C

∵ f(0)=0       ∴ C=π103

Now, f(1)=-153tan-1(35)+π103

⇒f(1)=153cot-1(35)=153tan-153

So, α=5:β=3    ∴ α2+β2=28