The integral ∫ (e^(3logₑ2x)+5e^(2logₑ2x))/(e^(4logₑx)+5e^(3logₑx)-7e^(2logₑx))dx, x > 0, is equal to : (where c is a constant of integration)

Solution

Maths sir · 4 steps

Turn each e^(n logₑ a) into aⁿ, then the top is 4 times the derivative of the bottom.

1. Remove the exponentials. Use e^(n logₑ a) = aⁿ. The numerator becomes (2x)³ + 5(2x)² = 8x³ + 20x². The denominator becomes x⁴ + 5x³ - 7x².
2. Factor and cancel. The numerator is 4x²(2x + 5) and the denominator is x²(x² + 5x - 7). Since x > 0, the x² cancels.

   ∫ 4(2x+5)/(x²+5x-7) dx = 4∫ (x²+5x-7)'/(x²+5x-7) dx

3. Substitute. Put t = x² + 5x - 7, so dt = (2x + 5) dx. The integral becomes 4∫ dt/t = 4 logₑ|t| + c.
4. So. The integral equals 4 logₑ|x² + 5x - 7| + c, which is option (b).

The idea to keep. When you see e^(n logₑ(·)), rewrite it as a power first. Then check if the numerator is a constant times the derivative of the denominator, because that integrates to a log.

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