Let I(x)=∫ √((x+7)/x) dx and I(9)=12+7 logₑ 7. If I(1)=α+7 logₑ(1+2√2), then α⁴ is equal to _________.

Solution

Maths sir · 4 steps

Put x = t² and the root becomes √(t²+7), a standard integral you already know.

1. Substitute. Let x = t², so dx = 2t dt and √((x+7)/x) = √(t²+7)/t. The t cancels, leaving I = 2∫ √(t²+7) dt.

   2∫ √(t²+7) dt = t√(t²+7) + 7 ln(t+√(t²+7)) + C

2. Back to x. With t = √x, this gives I(x) = √x(x+7) + 7 ln(√x+√(x+7)) + C.
3. Find C. I(9) = √144 + 7 ln(3+4) + C = 12 + 7 ln 7 + C. This matches the given value, so C = 0.
4. So. I(1) = √8 + 7 ln(1+2√2), so α = 2√2 and α⁴ = (√8)⁴ = 8² = 64.

The idea to keep. When you see √x under a root, try x = t². Then ∫√(t²+a²) dt = t/2√(t²+a²)+a²/2 ln(t+√(t²+a²)) finishes the job.

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