If ∫ (2eˣ+3e⁻ˣ)/(4eˣ+7e⁻ˣ)dx=1/14(ux+vlogₑ(4eˣ+7e⁻ˣ))+C, where C is a constant of integration, then u + v is equal to _____________.

Solution

Maths sir · 4 steps

Split the numerator into a part of the denominator plus a part of its derivative, then integrate each piece.

1. Set up the split. Let D = 4eˣ + 7e⁻ˣ, so D' = 4eˣ - 7e⁻ˣ. Write the numerator as 2eˣ + 3e⁻ˣ = A D + B D'.
2. Match coefficients. Comparing eˣ terms gives 4A + 4B = 2, so A + B = 1/2. Comparing e⁻ˣ terms gives 7A - 7B = 3, so A - B = 3/7. Solving, A = 13/28 and B = 1/28.

   ∫ (A D + B D')/D dx = 13/28x + 1/28 logₑ(4eˣ + 7e⁻ˣ) + C

3. Match the given form. Take 1/14 out: the result is 1/14(13/2x + 1/2 logₑ(4eˣ + 7e⁻ˣ)) + C. So u = 13/2 and v = 1/2.
4. So. u + v = 13/2 + 1/2 = 7.

The common slip: 14. Reading u and v off the 1/28 form gives u = 13, v = 1 and a total of 14. The bracket must be multiplied by 1/14, not 1/28, so both halve.

The idea to keep. For a ratio of exponential combinations, write numerator = A(denominator) + B(derivative of denominator). The A part gives Ax, the B part gives B log|D|.

Answer from the NTA final answer key. Solution by PrepWiser.