If ∫ (√(1+x²)+x)¹⁰/(√(1+x²)-x)⁹ dx=1/m((√(1+x²)+x)ⁿ(n√(1+x²)-x))+C where C is the constant of integration and m,n∈ N, then m+n is equal to _________ .

Solution

Maths sir · 5 steps

Since √(1+x²)-x is the reciprocal of √(1+x²)+x, the whole integrand becomes one power, t¹⁹.

1. Use the reciprocal. (√(1+x²)+x)(√(1+x²)-x) = 1+x²-x² = 1. So dividing by (√(1+x²)-x)⁹ is the same as multiplying by (√(1+x²)+x)⁹, and the integrand is (√(1+x²)+x)¹⁹.
2. Substitute. Let t = √(1+x²)+x, so 1/t = √(1+x²)-x. Subtracting gives x = 1/2(t-1/t), so dx = 1/2(1+1/t²)dt. Adding gives √(1+x²) = 1/2(t+1/t).

   ∫ t¹⁹·1/2(1+1/t²)dt = t²⁰/40+t¹⁸/36+C

3. Expand the given form. Using the expressions for √(1+x²) and x, tⁿ(n√(1+x²)-x) = 1/2[(n-1)tⁿ⁺¹+(n+1)tⁿ⁻¹]. Matching the top power t²⁰ needs n = 19, which gives 1/2(18t²⁰+20t¹⁸) = 9t²⁰+10t¹⁸.
4. Match coefficients. We need 9/m = 1/40 and 10/m = 1/36. Both give m = 360, so the two terms agree.
5. So. m = 360 and n = 19, so m+n = 379.

The idea to keep. Whenever you see √(1+x²)± x together, they are reciprocals. Put t = √(1+x²)+x, and then x and √(1+x²) are both simple expressions in t.

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