Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main202313 Apr 2023Morning ShiftMathematicsIndefinite IntegrationActual

∫ 0 ∞ 6 e 3 x + 6 e 2 x + 11 e x + 6 d x =

Options

  1. Alog e 32 27
  2. Blog e 512 81
  3. Clog e 256 81
  4. Dlog e 302 27

Correct answer

A. log e 32 27

Step-by-step solution

Let I = ∫ 0 ∞ 6 e 3 x + 6 e 2 x + 11 e x + 6 d x ⇒ I = ∫ 0 ∞ 6 e x + 1 e x + 2 e x + 3 d x Using partial fraction, we can write ⇒ I = 6 ∫ 0 ∞ 1 2 e x + 1 - 1 e x + 2 + 1 2 e x + 3 d x ⇒ I = 6 ∫ 0 ∞ 1 2 e - x + 1 - 1 1 + 2 e - x + 1 2 1 + 3 e - x e - x d x ⇒ I = 6 - 1 2 log e e - x + 1 + 1 2 log e 2 e - x + 1 - 1 6 log e 3 e - x + 1 0 ∞ ⇒ I = 6 0 - - 1 2 log e 2 + 1 2 log e 3 - 1 6 log e 4 ⇒ I = 3 log e 2 - 3 log e 3 + log e 4

Practice Indefinite Integration on Quantrex Academy →

More from Indefinite Integration

Let f(x) = ( 16x + 24 x^2 + 2x - 15 ) dx . If f(4) = 14 _e(3) and f(7) = _e(2^ 3^ ) , , N , then + is equal to: 2026Let f(x)= d x x^ ( 2 3 ) +2 x^ ( 1 2 ) be such that f(0)=-26+24 _ e (2) . If f(1)= a + b _ e (3) , where a , b Z , then a + b is equal to : 2026Let f(t)= ( 1- ( _ e t ) 1- ( _ e t ) ) d t, t>1 . If f (e^ / 2 )=-e^ / 2 and f (e^ / 4 )= e^ / 4 , then equals 2026Let I (x)= 3 d x (4 x+6) ( 4 x²+8 x+3 ) and I (0)= 3 4 +20 . If I ( 1 2 )= a 2 b + c , where a, b, c N , gcd (a, b)=1 , then a+b+c is equal to 2026Let f(x)= (2-x² ) e ^ x ( 1+x )(1-x)^ 3 / 2 ~d x . If f(0)=0 , then f ( 1 2 ) is equal to: 2026If ( x)^ -11 2 ( x)^ -5 2 d x=- p₁ q₁ ( x)^ 9 2 - p₂ q₂ ( x)^ 5 2 - p₃ q₃ ( x)^ 1 2 + p₄ q₄ ( x)^ -3 2 + C , where p_ i and q_ i are positive integers with gcd (p_ i , q_ i )=1 for 2026If aligned ( 1 x + 1 x^3 ) & ( [23] 3 x⁻²⁴+x⁻²⁶ ) d x & =- 3( +1) (3 x^ +x^ )^ +1 +C, x 0, aligned ( , , Z) , where C is the constant of integration, then + + is equal to ________ 2025If ( 1+x^2 +x )¹⁰ ( 1+x^2 -x )^9 d x= 1 m ( ( 1+x^2 +x )^n (n 1+x^2 -x ) )+C where C is the constant of integration and m, n N , then m + n is equal to 2025 Full Indefinite Integration list All JEE Main PYQs