Integrals of class of functions following a definite pattern can be found by the method of reduction and recursion. Reduction formulas make it possible to reduce an integral dependent on the index n > 0, called the order of the integral, to an integral of the same type with a smaller index. Integration by parts helps us to derive reduction formulas.
(i) If I n = 
then I n+1 + 
. 
I n is equal to
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Ans.
(i)
Sol. Using integration by parts
I n = 
+ 2n 
dx
= 
+ 2n 
dx
– 2na 2 
dx
whence I n+1 + 
. 
I n
= 
. 
(ii) Sol. I n,–m = 
– 
dx
= 
– 
I n–2,2–m
(iii)
Sol. Consider
u n+1 = 
dx
= 
dx
= 
dx – 
u n
= I n – 
u n , where ... (i)
I n = 
dx
= 
x n 2 
– 
nx n–1 2 
dx
= 
–
dx
aI n = x
n 
–na u n+1 – 2bn u n –nc u n–1
Putting this value in (i) we have
a u n+1 = x n 
–na u n+1 –2bn
u n – ncu n–1 –bu n
⇒ (n + 1) au n+1 + (2n +1) bu n + ncu n–1 = x n 
(iv)
Sol. I= 
x m (x 2m + x m +1) (2x 2m + 3x m +6) 1/m dx
= 
x m–1 (x 2m + x m + 1) (x m (2x 2m + 3x m + 6)) 1/m dx
= 
x m–1 (x 2m + x m +1) (2x 3m +3x 2m + 6x m ) 1/m dx
Put x m = t ⇒ mx m–1 dx = dt
∴ I = 
(t 2 + t+ 1) (2t 3 +3t 2 + 6t) 1/m dt
Put 2t 3 + 3t 2 + 6t = u m ⇒ (6t 2 + 6t + 6)
dt = mu m–1 dx
⇒ 
(t 2 + t+ 1) dt = 
u m–1 du
I = 
u m–1 u du = 
u m du = 
u m+1 = 
(2t
3 + 3t 2 + 6t 
= 
(2x 3m + 3x 2m + 6x m 
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