Explanation
Solution:
Step 1: Substitution
Let x2=t1 differentiate karne par 2xdx=−t21dt⟹dx=−2t3/2dt.
Lekin isse asan tarika hai:
t=x4−3x2⟹t2=x24−3x2=x24−3
Differentiating both sides:
2tdt=−x38dx⟹x3dx=−4tdt
Step 2: Integral ko simplify karna
Integral ko x ki powers mein rearrange karne par:
f(x)=∫x2(x23+4)⋅xx24−31dx=∫x3(x23+4)x24−31dx
Ab t ki values substitute karein:
f(x)=∫(43(t2+3)+4)⋅t−t/4dt=∫43t2+9+16−1/4dt
f(x)=−∫3t2+25dt=−31∫t2+(35)2dt
Step 3: Integration apply karna
Standard formula ∫x2+a2dx=a1tan−1(ax) use karne par:
f(x)=−31⋅5/31tan−1(5/3t)+C
f(x)=−531tan−1(5x34−3x2)+C
tan−1(θ1)=2π−tan−1(θ) ka upyog karke ise transform karein:
f(x)=531tan−1(34−3x25x)+C′
Step 4: Values nikalna
f(0)=0 diya hai, toh C′=0.
Ab x=1 rakhein:
f(1)=531tan−1(34−3(1)25(1))=531tan−1(35)
Humein diya hai f(1)=αβ1tan−1(βα). Compare karne par:
Step 5: Final Result
Answer: 28
