If (√1+x^2+x)^10(√1+x^2-x)^9dx=(1)/(m)((√1+x^2+x)^n(n√1+x^2-x))+C where C is the constant …

Question

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

Answer: 379

Accepted Answer

$\begin{gathered} \sqrt{1+{x}^{2}}+x=\sec 	heta +	an 	heta =t \ \sqrt{1+{x}^{2}}=\sec 	heta =\frac{{t}^{2}+1}{2t} \ x=	an 	heta =\frac{{t}^{2}-1}{2t} \end{gathered}$

The given expression becomes

$\frac{1}{m}{t}^{2}(n\cdot \frac{{t}^{2}+1}{2t}-\frac{{t}^{2}-1}{2t})=\frac{{t}^{n-1}}{2m}((n-1){t}^{2}+n+1)$

By compare

$\begin{gathered} n=19 \ m=360 \ 	herefore \, n+m=379 \end{gathered}$

Solutions · 1