Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice

The value of the integral ∫ e x 2 + 1 x 2 x 2 - 1 x + 1 d x is equal to (where C is the constant of integration)

Options

  1. Ae x 2 + 1 x + C
  2. Bx 2 e x 2 + 1 x + C
  3. Cx e x 2 + 1 x + C
  4. Dx ⋅ e x 2 + C

Correct answer

C. x e x 2 + 1 x + C

Step-by-step solution

Ι = ∫ e x 2 + 1 x ⏟ Ι ⋅ 1 ⏟ Ι Ι d x + ∫ e x 2 + 1 x 2 x - 1 x 2 ⋅ x d x Using integration by parts, we get, Ι = e x 2 + 1 x · x - ∫ e x 2 + 1 x 2 x - 1 x 2 ⋅ x d x + ∫ e x 2 + 1 x 2 x - 1 x 2 ⋅ x d x = x e x 2 + 1 x + C

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 NTA Abhyas JEE Main PYQs