Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

 ❓ Question Evaluate the integral: ∫ ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3   d x = − α 3 ( α + 1 ) ( 3 x β + x γ ) α / ( α + 1 ) + C , where C C  is the constant of integration. Find α + β + γ \alpha + \beta + \gamma . 🖼️ Question Image ✍️ Short Solution Factor inside the cube root: 3 x − 24 + x − 26 = x − 26 ( 3 x 2 + 1 ) Then, cube root: 3 x − 24 + x − 26 3 = x − 26 ( 3 x 2 + 1 ) 3 = x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . Simplify the squared bracket: ( 1 x + 1 x 3 ) = x 2 + 1 x 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) = \frac{x^2+1}{x^3}. Raise to first power (as given), then multiply with cube root: ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3 = x 2 + 1 x 3 ⋅ x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 = ( x 2 + 1 ) x − 3 − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) \sqrt[3]{3x^{-24} + x^{-26}} = \frac{x^2+1}{x^3} \cdot x^{-26/3} (3x^2 +1)^{1/3} = (x^2+1) x^{-3-26/3} (3x^2+1)^{1/3}. Compute exponent: − 3 − 26 / 3 = − 35 / 3 -3 - 26/3 = -35/3 . So: (...

  Question Let y = y ( x ) y = y(x) be the solution of the differential equation ( x 2 + 1 ) y ′ − 2 x y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x , with y ( 0 ) = 1 y(0) = 1 . Then evaluate: ∫ − 3 3 y ( x )   d x Question Image Short Solution Rewrite the ODE in standard linear form: y ′ − 2 x x 2 + 1 y = ( x 4 + 2 x 2 + 1 ) cos ⁡ x x 2 + 1​ Find the integrating factor : μ ( x ) = e ∫ − 2 x x 2 + 1 d x = e − ln ⁡ ( x 2 + 1 ) = 1 x 2 + 1​ Multiply the equation by μ ( x ) \mu(x) : d d x ( y x 2 + 1 ) = cos ⁡ x Integrate: y x 2 + 1 = sin ⁡ x + C Apply y ( 0 ) = 1 y(0)=1 : 1 1 = 0 + C    ⟹    C = 1 Hence: y ( x ) = ( x 2 + 1 ) ( sin ⁡ x + 1 ) Compute the definite integral: I = ∫ − 3 3 y ( x )   d x = ∫ − 3 3 ( x 2 + 1 ) ( sin ⁡ x + 1 ) d x Image Solution Conclusion Break the integral: I = ∫ − 3 3 ( x 2 + 1 ) sin ⁡ x   d x + ∫ − 3 3 ( x 2 + 1 ) d x First term: ( x 2 + 1 ) sin ⁡ x (x^{2}+1)\sin x  is an odd function (because x 2 + 1 x^{2}+1  is eve...