If $\int \frac{\left(\sqrt{1+\mathrm{x}^{2}}+\mathrm{x}\right)^{10}}{\left(\sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)^{9}} \mathrm{dx}=\frac{1}{\mathrm{~m}}\left(\left(\sqrt{1+\mathrm{x}^{2}}+\mathrm{x}\right)^{\mathrm{n}}\left(\mathrm{n} \sqrt{1+\mathrm{x}^{2}}-\mathrm{x}\right)\right)+\mathrm{C}$ where C is the constant of integration and $\mathrm{m}, \mathrm{n} \in \mathrm{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Indefinite Integrals

\int \frac{( 1 + x ^{2} + x ) ^{10}}{( 1 + x ^{2} - x ) ^{9}} dx = \frac{1}{m} ((1 + x^{2} + x)^{n} (n 1 + x^{2} - x)) + C

where C is the constant of integration and

m, n \in N

, then

m + n

is equal to

Answer: 379

Since

1 + x^{2} - x = \frac{1}{t}

for

t = 1 + x^{2} + x

, the integrand reduces to

t^{19}

, and

d x = \frac{1}{2} (1 + t^{- 2}) d t

. Then

\int \frac{1}{2} (t^{19} + t^{17}) d t = \frac{t ^{18}}{360} (9 t^{2} + 10) = \frac{1}{360} (1 + x^{2} + x)^{19} (19 1 + x^{2} - x) + C

. Thus

m = 360, n = 19

and

m + n = 379

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Integer answer