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Integrals · Question 40

JEE Main 2025 (Online) 4th April Evening ShiftNumericalHard📈 Topic in 23% papers+4
1

If ( 1+x2+x )10 ( 1+x2-x )9 dx= 1m ( ( 1+x2+x )n (n 1+x2-x ) )+C where C is the constant of integration and m, n N, then m+n is equal to .

Answer

379

Solution

Let t= 1+x2+x. Then 1+x2-x= 1t

1+x2= 12 (t+ 1t ), x= 12 (t- 1t )

dtdx= x 1+x2+1= t 1+x2 dx= 12 (1+ 1t2 )dt

I= t10 t9 12 (1+ 1t2 )dt= 12 (t19+t17 )dt

I= t2040+ t1836+C

Right side: n 1+x2-x= 12 [(n-1)t+ n+1t ]

1mtn (n 1+x2-x )= (n-1)tn+1+(n+1)tn-12m

Comparing: n+1=20 n=19

n-12m= 140 182m= 140 m=360

Check: n+12m= 20720= 136 (matches)

m+n=360+19=379

Answer: 379

JEE Main 2025 (Online) 4th April Evening Shift

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