Let f(x)=∫ 2x/(x²+1)(x²+3)dx. If f(3)=1/2(logₑ5-logₑ6), then f(4) is equal to

Solution

Maths sir · 4 steps

See 2x dx sitting on top? Put t = x² and the integral becomes plain partial fractions.

1. Substitute. Let t = x², so dt = 2x dx. The integral becomes ∫ dt/(t+1)(t+3).
2. Partial fractions. Write 1/(t+1)(t+3) = 1/2(1/(t+1) - 1/(t+3)). Integrate each term. Both t+1 and t+3 are positive, so no modulus is needed.

   f(x) = 1/2 ln((x²+1)/(x²+3)) + C

3. Find C. At x = 3: f(3) = 1/2 ln 10/12 + C = 1/2(ln 5 - ln 6) + C. This already equals the given value, so C = 0.
4. So. f(4) = 1/2 ln 17/19 = 1/2(logₑ 17 - logₑ 19), which is option (d).

The idea to keep. When the numerator is the derivative of the inner expression, substitute it first. Then use the given value to fix the constant before you find the new value.

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